CF1076E：Vasya and a Tree（DFS&差分）

Posted 2021-01-23 hua-dong

Vasya has a tree consisting of n n vertices with root in vertex 1 1 . At first all vertices has 0 0 written on it.

Let d(i,j) d(i,j) be the distance between vertices i i and j j , i.e. number of edges in the shortest path from i i to j j . Also, let‘s denote k k -subtree of vertex x x — set of vertices y y such that next two conditions are met:

• x x is the ancestor of y y (each vertex is the ancestor of itself);
• d(x,y)≤k d(x,y)≤k .

Vasya needs you to process m m queries. The i i -th query is a triple v i  vi , d i  di and x i  xi . For each query Vasya adds value x i  xi to each vertex from d i  di -subtree of v i  vi .

Report to Vasya all values, written on vertices of the tree after processing all queries.

Input

The first line contains single integer n n (1≤n≤3⋅10 5  1≤n≤3⋅105 ) — number of vertices in the tree.

Each of next n−1 n−1 lines contains two integers x x and y y (1≤x,y≤n 1≤x,y≤n ) — edge between vertices x x and y y . It is guarantied that given graph is a tree.

Next line contains single integer m m (1≤m≤3⋅10 5  1≤m≤3⋅105 ) — number of queries.

Each of next m m lines contains three integers v i  vi , d i  di , x i  xi (1≤v i ≤n 1≤vi≤n , 0≤d i ≤10 9  0≤di≤109 , 1≤x i ≤10 9  1≤xi≤109 ) — description of the i i -th query.

Output

Print n n integers. The i i -th integers is the value, written in the i i -th vertex after processing all queries.

Examples

Input

5
1 2
1 3
2 4
2 5
3
1 1 1
2 0 10
4 10 100

Output

1 11 1 100 0 

Input

5
2 3
2 1
5 4
3 4
5
2 0 4
3 10 1
1 2 3
2 3 10
1 1 7

Output

10 24 14 11 11 

Note

In the first exapmle initial values in vertices are 0,0,0,0,0 0,0,0,0,0 . After the first query values will be equal to 1,1,1,0,0 1,1,1,0,0 . After the second query values will be equal to 1,11,1,0,0 1,11,1,0,0 . After the third query values will be equal to 1,11,1,100,0 1,11,1,100,0

 

题意：给定一棵大小为N个树，Q次操作，每次给出三元组(u,d,x)表示给u为根的子树，距离u不超过d的点加值x。

思路：对于每个操作，我们在u处加x，在dep[u+d+1]处减去x。只需要传递一个数组，代表在深度为多少的时候减去多少即可，由于是DFS，满足操作都是在子树里的。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
using namespace std;
const int maxn=1000010;
int dep[maxn],N,Laxt[maxn],Next[maxn],To[maxn],cnt;
int laxt2[maxn],next2[maxn],D[maxn],X[maxn],tot; ll ans[maxn];
void add(int u,int v){
    Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v;
}
void add2(int u,int d,int x){
    next2[++tot]=laxt2[u]; laxt2[u]=tot; D[tot]=d; X[tot]=x;
}
void dfs(int u,int f,ll sum,ll *mp)
{
    dep[u]=dep[f]+1; sum-=mp[dep[u]];
    for(int i=laxt2[u];i;i=next2[i]){
        sum+=X[i];if(dep[u]+D[i]+1<=N) mp[dep[u]+D[i]+1]+=X[i];
    }
    ans[u]=sum;
    for(int i=Laxt[u];i;i=Next[i])
      if(To[i]!=f) dfs(To[i],u,sum,mp);
    for(int i=laxt2[u];i;i=next2[i]){
        sum-=X[i];if(dep[u]+D[i]+1<=N) mp[dep[u]+D[i]+1]-=X[i];
    }
}
ll mp[maxn];
int main()
{
    int u,v,x,Q; scanf("%d",&N);
    rep(i,1,N-1) {
        scanf("%d%d",&u,&v);
        add(u,v); add(v,u);
    }
    scanf("%d",&Q);
    rep(i,1,Q) {
        scanf("%d%d%d",&u,&v,&x);
        add2(u,v,x);
    }

    dfs(1,0,0LL,mp);
    rep(i,1,N) printf("%lld ",ans[i]);
    return 0;
}