小Y有一棵n个节点的树，每条边都有正的边权。

小J有q个询问，每次小J会删掉这个树中的k条边，这棵树被分成k+1个连通块。小J想知道每个连通块中最远点对距离的和。

这里的询问是互相独立的，即每次都是在小Y的原树上进行操作。

共q行，每行一个整数表示询问的答案。

5
1 2 2
2 3 3
2 4 4
4 5 2
3
4 1 2 3 4
1 4
2 2 3

0
7
4

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 100010
#define maxm 200010
#define maxlog 20
#define oo 2147483647
#define LL long long
int n, m, head[maxn], Next[maxm], To[maxm], dist[maxm];
struct Edge {
	int a, b, c;
	Edge() {}
	Edge(int _1, int _2, int _3): a(_1), b(_2), c(_3) {}
} es[maxn];

void AddEdge(int a, int b, int c) {
	To[++m] = b; dist[m] = c; Next[m] = head[a]; head[a] = m;
	swap(a, b);
	To[++m] = b; dist[m] = c; Next[m] = head[a]; head[a] = m;
	return ;
}

int dep[maxn], list[maxm], cl, pos[maxn], ord[maxn], clo, ordr[maxn], Pos[maxn];
LL Dep[maxn];
void build(int u, int fa) {
	list[++cl] = u; pos[u] = cl; ord[u] = ++clo; Pos[clo] = u;
	for(int e = head[u]; e; e = Next[e]) if(To[e] != fa) {
		dep[To[e]] = dep[u] + 1;
		Dep[To[e]] = Dep[u] + dist[e];
		build(To[e], u);
		list[++cl] = u;
	}
	ordr[u] = clo;
	return ;
}
int Lca[maxlog][maxm], Log[maxm];
void rmq_init() {
	Log[1] = 0;
	for(int i = 2; i <= cl; i++) Log[i] = Log[i>>1] + 1;
	for(int i = 1; i <= cl; i++) Lca[0][i] = list[i];
	for(int j = 1; (1 << j) <= cl; j++)
		for(int i = 1; i + (1 << j) - 1 <= cl; i++) {
			int a = Lca[j-1][i], b = Lca[j-1][i+(1<<j-1)];
			if(dep[a] < dep[b]) Lca[j][i] = a;
			else Lca[j][i] = b;
		}
	return ;
}
int lca(int a, int b) {
	int l = min(pos[a], pos[b]), r = max(pos[a], pos[b]);
	int t = Log[r-l+1];
	a = Lca[t][l]; b = Lca[t][r-(1<<t)+1];
	return dep[a] < dep[b] ? a : b;
}
LL calc(int a, int b) {
	return Dep[a] + Dep[b] - (Dep[lca(a,b)] << 1);
}

struct Node {
	bool hasv;
	int A, B; LL Len;
	Node() {}
	Node(int _1, int _2, LL _3, bool _4): A(_1), B(_2), Len(_3), hasv(_4) {}
} ns[maxn<<2];
void maintain(Node& o, Node lc, Node rc) {
	o.Len = -1;
	if(!lc.hasv && !rc.hasv){ o.Len = o.hasv = 0; return ; }
	if(!lc.hasv) {
		o = rc;
		return ;
	}
	if(!rc.hasv) {
		o = lc;
		return ;
	}
	o.hasv = 1;
	if(o.Len < lc.Len) o.Len = lc.Len, o.A = lc.A, o.B = lc.B;
	if(o.Len < rc.Len) o.Len = rc.Len, o.A = rc.A, o.B = rc.B;
	LL d = calc(lc.A, rc.A);
	if(o.Len < d) o.Len = d, o.A = lc.A, o.B = rc.A;
	d = calc(lc.A, rc.B);
	if(o.Len < d) o.Len = d, o.A = lc.A, o.B = rc.B;
	d = calc(lc.B, rc.A);
	if(o.Len < d) o.Len = d, o.A = lc.B, o.B = rc.A;
	d = calc(lc.B, rc.B);
	if(o.Len < d) o.Len = d, o.A = lc.B, o.B = rc.B;
	return ;
}
void build(int L, int R, int o) {
	if(L == R) ns[o] = Node(Pos[L], Pos[R], 0, 1);
	else {
		int M = L + R >> 1, lc = o << 1, rc = lc | 1;
		build(L, M, lc); build(M+1, R, rc);
		maintain(ns[o], ns[lc], ns[rc]);
		ns[o].hasv = 1;
	}
//	printf("seg[%d, %d]: %d %d %lld\n", L, R, ns[o].A, ns[o].B, ns[o].Len);
	return ;
}
void update(int L, int R, int o, int ql, int qr, bool v) {
	if(ql <= L && R <= qr) ns[o].hasv = v;
	else {
		int M = L + R >> 1, lc = o << 1, rc = lc | 1;
		if(ql <= M) update(L, M, lc, ql, qr, v);
		if(qr > M) update(M+1, R, rc, ql, qr, v);
		maintain(ns[o], ns[lc], ns[rc]);
	}
	return ;
}
Node query(int L, int R, int o, int ql, int qr) {
	if(ql <= L && R <= qr) return ns[o];
	int M = L + R >> 1, lc = o << 1, rc = lc | 1;
	Node ans(-1, -1, -1, 0);
	if(ql <= M) {
		Node tmp = query(L, M, lc, ql, qr), tt(0, 0, -1, 1);
		maintain(tt, tmp, ans);
		if(tmp.hasv) ans = tt;
	}
	if(qr > M) {
		Node tmp = query(M+1, R, rc, ql, qr), tt(0, 0, -1, 1);
		maintain(tt, tmp, ans);
		if(tmp.hasv) ans = tt;
	}
//	printf("[%d, %d] %d %d %d %lld(%d)\nans: %lld\n", L, R, o, ql, qr, ns[o].Len, ns[o].hasv, ans.Len);
	return ans;
}

bool cmp(int a, int b) { return pos[a] < pos[b]; }
int psi[maxn], cpi, ps[maxn], cp, vis[maxn];
bool flg[maxn];

struct Vtree {
	int m, head[maxn], Next[maxm], To[maxm];
	LL dist[maxm], ans;
	void init() {
		ans = m = 0;
		return ;
	}
	void AddEdge(int a, int b, LL c) {
//		printf("Add2: %d %d %lld\n", a, b, c);
		To[++m] = b; dist[m] = c; Next[m] = head[a]; head[a] = m;
		swap(a, b);
		To[++m] = b; dist[m] = c; Next[m] = head[a]; head[a] = m;
		return ;
	}
	void dfs(int u, int fa) {
		for(int e = head[u]; e; e = Next[e]) if(To[e] != fa)
			dfs(To[e], u);
		if(flg[u]) {
			Node tmp = query(1, n, 1, ord[u], ordr[u]); flg[u] = 0;
			if(tmp.hasv) ans += tmp.Len;
//			printf("at %d tmp: %lld [%d, %d]\n", u, tmp.Len, ord[u], ordr[u]);
			update(1, n, 1, ord[u], ordr[u], 0);
		}
		return ;
	}
	void dfs2(int u, int fa) {
		for(int e = head[u]; e; e = Next[e]) if(To[e] != fa)
			dfs2(To[e], u);
		update(1, n, 1, ord[u], ordr[u], 1);
		head[u] = 0;
		return ;
	}
} vt;

int main() {
	n = read();
	for(int i = 1; i < n; i++) {
		int a = read(), b = read(), c = read();
		AddEdge(a, b, c);
		es[i] = Edge(a, b, c);
	}
	build(1, 0);
	rmq_init();
//	for(int i = 1; i <= cl; i++) printf("%d%c", Lca[0][i], i < cl ? ‘ ‘ : ‘\n‘);
//	for(int i = 1; i <= n; i++) printf("%d%c", pos[i], i < n ? ‘ ‘ : ‘\n‘);
	build(1, n, 1);
	int q = read();
//	for(int i = 1; i <= n; i++) printf("%d%c", ord[i], i < n ? ‘ ‘ : ‘\n‘);
	while(q--) {
		int cpi = read(); cp = 0;
		for(int i = 1; i <= cpi; i++) {
			int e = read(), u = dep[es[e].a] < dep[es[e].b] ? es[e].b : es[e].a;
			ps[++cp] = psi[i] = u;
			vis[ps[cp]] = q + 1;
			flg[ps[cp]] = 1;
		}
		if(vis[1] != q + 1) ps[++cp] = 1, flg[1] = 1;
		sort(psi + 1, psi + cpi + 1, cmp);
		for(int i = 1; i < cpi; i++) {
			int c = lca(psi[i], psi[i+1]);
			if(vis[c] != q + 1) vis[c] = q + 1, ps[++cp] = c;
		}
		sort(ps + 1, ps + cp + 1, cmp);
//		printf("mother_fuck: "); for(int i = 1; i <= cp; i++) printf("%d%c", ps[i], i < cp ? ‘ ‘ : ‘\n‘);
		vt.init();
		for(int i = 1; i < cp; i++) {
			int a = ps[i], b = ps[i+1], c = lca(a, b);
			vt.AddEdge(b, c, Dep[b] - Dep[c]);
		}
		vt.dfs(1, 0);
		vt.dfs2(1, 0);
		printf("%lld\n", vt.ans);
	}
	
	return 0;
}
/*
13
1 2 1
1 3 1
2 4 1
2 5 1
2 6 1
3 7 1
3 8 1
3 9 1
7 12 1
12 13 1
9 10 1
10 11 1
*/