// bzoj 1036#include<bits/stdc++.h>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define FF(i,a,b) for (int i=a;i<=b;i++)#define F(i,b,a)  for (int i=b;i>=a;i--)#define mes(a,b)  memset(a,b,sizeof(a))#define INF 0x3f3f3f3ftypedef long long ll;const int N = 3e4+10;struct Edge{int to, next;}e[N];struct Point{int mx, sum;}t[N<<2];int n, sz, Size[N], dep[N], w[N], head[N], pos[N], tot, bl[N], fa[N];void Addedge(int u, int v) { e[tot].to=v, e[tot].next=head[u], head[u]=tot++; }void Init(){    mes(head,-1);    scanf("%d", &n);    int u, v;    FF(i,1,n-1) {        scanf("%d %d", &u, &v);        Addedge(u, v);  Addedge(v, u);    }}void dfs1(int u)   //搜索出各个子树的Size{    Size[u]=1;    for(int i=head[u]; i!=-1; i=e[i].next) if(e[i].to!=fa[u]) {        int v=e[i].to;        fa[v]=u, dep[v]=dep[u]+1;        dfs1(v);        Size[u]+=Size[v];    }}void dfs2(int u, int chain)    //构建树链{    sz++, pos[u]=sz, bl[u]=chain;   //pos[]分配u结点在线段树中的编号，chain是一条链上头结点的标号    int mx=0;    for(int i=head[u]; i!=-1; i=e[i].next) {        int v=e[i].to;        if(dep[v]>dep[u] && Size[v]>Size[mx]) mx=v;    }    if(mx==0) return ;    dfs2(mx, chain);    for(int i=head[u]; i!=-1; i=e[i].next) {        int v=e[i].to;        if(dep[v]>dep[u] && mx!=v) dfs2(v, v);    }}void update(int ro, int l, int r, int x, int y)    //线段树单点更新{    if(l==r) { t[ro].mx=t[ro].sum=y; return ; }    int mid=(l+r)>>1;    if(x<=mid) update(ro<<1, l, mid, x, y);    else update(ro<<1|1, mid+1, r, x, y);    t[ro].mx=max(t[ro<<1].mx, t[ro<<1|1].mx);    t[ro].sum=t[ro<<1].sum+t[ro<<1|1].sum;}int querymx(int ro, int l, int r, int x, int y)     //线段树区间求最大值{    if(l==x && r==y) return t[ro].mx;    int mid=(l+r)>>1;    if(y<=mid) return querymx(ro<<1, l, mid, x, y);    else if(mid<x) return querymx(ro<<1|1, mid+1, r, x, y);    else return max(querymx(ro<<1, l, mid, x, mid), querymx(ro<<1|1, mid+1, r, mid+1, y));}int solvemx(int x, int y){    int mx=-INF;    while(bl[x]!=bl[y]) {        if(dep[bl[x]]<dep[bl[y]]) swap(x, y);        mx=max(mx, querymx(1, 1, sz, pos[bl[x]], pos[x]));        x=fa[bl[x]];    }    if(pos[x]>pos[y]) swap(x, y);    mx=max(mx, querymx(1, 1, sz, pos[x], pos[y]));    return mx;}int querysum(int ro, int l, int r, int x, int y)       //线段树区间求和{    if(l==x && r==y) return t[ro].sum;    int mid=(l+r)>>1;    if(y<=mid) return querysum(ro<<1, l, mid, x, y);    else if(mid<x) return querysum(ro<<1|1, mid+1, r, x, y);    else return querysum(ro<<1, l, mid, x, mid)+querysum(ro<<1|1, mid+1, r, mid+1, y);}int solvesum(int x, int y){    int sum=0;    while(bl[x]!=bl[y]) {        //不在一条重链上时，就一边修改，一边往同一条重链上靠        if(dep[bl[x]]<dep[bl[y]]) swap(x, y);        sum+=querysum(1, 1, sz, pos[bl[x]], pos[x]);        x=fa[bl[x]];    }    if(pos[x]>pos[y]) swap(x, y);    sum+=querysum(1, 1, sz, pos[x], pos[y]);    return sum;}void solve(){    FF(i,1,n) scanf("%d", &w[i]), update(1, 1, sz, pos[i], w[i]);    int q, x, y;    char str[10];    scanf("%d", &q);    FF(i,1,q) {        scanf("%s %d %d", str, &x, &y);        if(str[0]==‘C‘) w[x]=y, update(1, 1, sz, pos[x], y);        else if(str[1]==‘M‘) printf("%d\n", solvemx(x, y));        else printf("%d\n", solvesum(x, y));    }}int main(){    Init();    dfs1(1);    dfs2(1, 1);    solve();    return 0;}