/*51 nod 1766 树上的最远点对(线段树+lca)problem:n个点被n-1条边连接成了一颗树，给出a~b和c~d两个区间，表示点的标号请你求出两个区间内各选一点之间的最大距离，即你需要求出max｛dis(i,j) |a<=i<=b,c<=j<=d｝solve:最开始想的是树链剖分,结果发现是区间[a,b]. 看成链了...看题解说的是最远点有合并的性质.  就是[a,b]的最远点ta,tb和[b+1,c]的最远点tc,td这四个点中距离最远的就是[a,c]的最远点..  证明并没有怎么看懂- -如果用线段树和并的话，需要快速求两点之间的距离.  可以用st快速求lca来解决,预处理便能得到O(1)的.然后就是线段树合并时处理下.http://blog.csdn.net/rzo_kqp_orz/article/details/52280811hhh-2016/09/15-21:16:26*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <set>#include <map>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfs(a) scanf("%s",a)#define scanfl(a) scanf("%I64d",&a)#define scanfd(a) scanf("%lf",&a)#define key_val ch[ch[root][1]][0]#define eps 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1e9+7;const int MAXN = 100010;const double PI = acos(-1.0);template<class T> void read(T&num){    char CH;    bool F=false;    for(CH=getchar(); CH<‘0‘||CH>‘9‘; F= CH==‘-‘,CH=getchar());    for(num=0; CH>=‘0‘&&CH<=‘9‘; num=num*10+CH-‘0‘,CH=getchar());    F && (num=-num);}int stk[70], tp;template<class T> inline void print(T p){    if(!p)    {        puts("0");        return;    }    while(p) stk[++ tp] = p%10, p/=10;    while(tp) putchar(stk[tp--] + ‘0‘);    putchar(‘\n‘);}int rmq[2*MAXN];struct ST{    int mm[2*MAXN];    int dp[2*MAXN][20];    void init(int n)    {        mm[0] = -1;        for(int i = 1; i <= n; i++)        {            mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];            dp[i][0] = i;        }        for(int j = 1; j <= mm[n]; j++)            for(int i = 1; i + (1<<j) - 1 <= n; i++)                dp[i][j] = rmq[dp[i][j-1]] < rmq[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];    }    int query(int a,int b)    {        if(a > b)swap(a,b);        int k = mm[b-a+1];        return rmq[dp[a][k]] <= rmq[dp[b-(1<<k)+1][k]]?dp[a][k]:dp[b-(1<<k)+1][k];    }};struct Edge{    int to,next;    ll w;};Edge edge[MAXN*2];int tot,head[MAXN];int F[MAXN*2];int P[MAXN];int cnt;ll dis[MAXN];ST st;void ini(){    tot = 0;    memset(head,-1,sizeof(head));}void add_edge(int u,int v,ll w){    edge[tot].to = v;    edge[tot].w = w;    edge[tot].next = head[u];    head[u] = tot++;}void dfs(int u,int pre,int dep){    F[++cnt] = u;    rmq[cnt] = dep;    P[u] = cnt;    for(int i = head[u]; i != -1; i = edge[i].next)    {        int v = edge[i].to;        if(v == pre)continue;        dis[v] = dis[u] + edge[i].w;        dfs(v,u,dep+1);        F[++cnt] = u;        rmq[cnt] = dep;    }}void LCA_init(int root,int node_num){    cnt = 0;    dfs(root,root,0);    st.init(2*node_num-1);}int query_lca(int u,int v){    return F[st.query(P[u],P[v])];}ll distan(int a,int b){    int lca = query_lca(a,b);    return dis[a]+dis[b]-dis[lca]-dis[lca];}struct node{    int l,r;    int s,t;    ll len;} tree[MAXN << 2];void cal(int i,int a,int b){    ll len = distan(a,b);    if(tree[i].len < len)    {        tree[i].len = len;        tree[i].s = a;        tree[i].t = b;    }}void push_up(int i){    cal(i,tree[lson].s,tree[rson].s);    cal(i,tree[lson].s,tree[rson].t);    cal(i,tree[lson].t,tree[rson].s);    cal(i,tree[lson].t,tree[rson].t);    cal(i,tree[lson].s,tree[lson].t);    cal(i,tree[rson].s,tree[rson].t);}void build(int i,int l,int r){    tree[i].l = l,tree[i].r = r;    tree[i].len = 0;    tree[i].s = tree[i].t = 0;    if(l == r)    {        tree[i].s = tree[i].t = l;        tree[i].len = 0;        return;    }    int mid = (tree[i].l + tree[i].r) >> 1;    build(lson,l,mid);    build(rson,mid+1,r);    push_up(i);}int from,to;void solve(int a,int b,ll &len){    if(distan(a,b) > len)    {        len = distan(a,b);        from = a,to = b;    }}void query(int i,int l,int r,int &ta,int &tb){    ta = tb = -1;    if(tree[i].l >= l && tree[i].r <= r)    {        ta = tree[i].s ;        tb = tree[i].t;        return ;    }    int mid = (tree[i].l + tree[i].r) >> 1;    int ls,lt,rs,rt;    ls = lt = rs = rt= -1;    if(r <= mid)        query(lson,l,r,ta,tb);    else if(l > mid)        query(rson,l,r,ta,tb);    else    {        ll tans = -1;        query(lson,l,mid,ls,lt);        query(rson,mid+1,r,rs,rt);        solve(ls,rt,tans);        solve(ls,rs,tans);        solve(lt,rt,tans);        solve(lt,rs,tans);        solve(ls,lt,tans);        solve(rs,rt,tans);        ta = from ,tb = to;    }    push_up(i);}int main(){//    freopen("in.txt","r",stdin);    int n;    int u,v,w;    read(n);    ini();    for(int i = 1; i < n; i++)    {        read(u),read(v),read(w);        add_edge(u,v,w);        add_edge(v,u,w);    }    dis[1] = 0;    LCA_init(1,n);    build(1,1,n);    int a,b,m;    int max1,min1,max2,min2;    read(m);    for(int i = 1; i <= m; i++)    {        ll ans = 0;        read(u),read(v);        read(a),read(b);        query(1,u,v,max1,min1);        query(1,a,b,max2,min2);//            cout << max1 << max2 << min1<<min2 <<endl;        ans = max(ans,distan(max1,max2));        ans = max(ans,distan(max1,min2));        ans = max(ans,distan(min1,max2));        ans = max(ans,distan(min1,min2));        printf("%I64d\n",ans);    }    return 0;}