Problem:
Score : 400 points

Problem Statement

Given is a rooted tree with N vertices numbered 1 to N.
The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i.

Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0.

Now, the following Q operations will be performed:

Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j.

Find the value of the counter on each vertex after all operations.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq Q \leq 2 \times 10^5

1 \leq a_i < b_i \leq N

1 \leq p_j \leq N

1 \leq x_j \leq 10^4

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N Q
a_1 b_1
:
a_{N-1} b_{N-1}
p_1 x_1
:
p_Q x_Q

Output

Print the values of the counters on Vertex 1, 2, \ldots, N after all operations, in this order, with spaces in between.

Sample Input 1

4 3
1 2
2 3
2 4
2 10
1 100
3 1

Sample Output 1

100 110 111 110

The tree in this input is as follows:

Each operation changes the values of the counters on the vertices as follows:

Operation 1: Increment by 10 the counter on every vertex contained in the subtree rooted at Vertex 2, that is, Vertex 2, 3, 4. The values of the counters on Vertex 1, 2, 3, 4 are now 0, 10, 10, 10, respectively.

Operation 2: Increment by 100 the counter on every vertex contained in the subtree rooted at Vertex 1, that is, Vertex 1, 2, 3, 4. The values of the counters on Vertex 1, 2, 3, 4 are now 100, 110, 110, 110, respectively.

Operation 3: Increment by 1 the counter on every vertex contained in the subtree rooted at Vertex 3, that is, Vertex 3. The values of the counters on Vertex 1, 2, 3, 4 are now 100, 110, 111, 110, respectively.

Sample Input 2

6 2
1 2
1 3
2 4
3 6
2 5
1 10
1 10

Sample Output 2

20 20 20 20 20 20

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
int ans[200000];
int n, q;
int a[200000], b[200000], p[200000], x[200000];
int *tree[200000];
int cnt[200000];
int cnts[200000];

int DFS(int now_v, int before_v) {//v:現在の頂点　bv:前にいた頂点
    int i, new_v;
    
	for(i = 0; i < cnt[now_v]; i++){//iをvの個数分だけループ
		new_v = tree[now_v][i];//お隣の頂点
		if (before_v != new_v){//逆流を防ぐため，自分が前にいたところには行かない
			ans[new_v] += ans[now_v];
			DFS(new_v, now_v);//現在の自分の位置を，前にいたところに変え，お隣の頂点に移動
		}
	}
	return 0;
}

int main(void){
    scanf("%d%d", &n, &q);
    int i, j;
    for(i = 0; i < n; i++){
        cnt[i] = 0;
        ans[i] = 0;
        cnts[i] = 0;
    }
    
    for(i = 0; i < n - 1; i++){
        scanf("%d%d", &a[i], &b[i]);
        a[i]--;
        b[i]--;
        cnt[a[i]]++;
        cnt[b[i]]++;
    }
    for(i = 0; i < q; i++){
        scanf("%d%d", &p[i], &x[i]);
        p[i]--;
    }

    //普通にメモリ確保すると足りなくなるので，可変長配列で確保する
    for(i = 0; i < n; i++){
        tree[i] = (int *)malloc(sizeof(int)*(cnt[i] + 1));
        if(tree[i] == NULL){
            return 0;
        }
    }

    for(i = 0; i < n - 1; i++){
        tree[a[i]][cnts[a[i]]] = b[i];
        tree[b[i]][cnts[b[i]]] = a[i];
        cnts[a[i]]++;
        cnts[b[i]]++;
    }

    for(i = 0; i < q; i++) {
		ans[p[i]] += x[i];
	}

	DFS(0, -1);

    for(i = 0; i < n; i++){
        printf("%d", ans[i]);
        if(i == n - 1){
            break;
        }
        printf(" ");
    }
    printf("\n");

    return 0;
}