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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

#define WHITE 0
#define GRAY 1
#define BLACK 2

//節点uからでる枝の構造体の定義
typedef struct slobj_{
    struct slobj_ *next;
    int v; //uから枝が伸びている節点v
}* slobj;

//slobj型のリストの定義
typedef struct{
    slobj head;
    slobj tail;
}* slist;

//グラフの構造体の定義
typedef struct{
    int n; //節点数
    slist* E; //枝リストの配列E[0]~E[n-1],未ソート単方向
}* graph;

//vにnを持つslobj型のリスト要素のメモリ確保+作成する関数
slobj slobj_new(int n){
    slobj p;
    NEW(p,1);
    p->v=n;
    p->next=NULL;
    return p;
}

//slist型のメモリを1個確保し、空リスト1つを作る関数
slist slist_new(void){
    slist L;
    NEW(L,1);
    L->head=NULL;
    L->tail=NULL;
    return L;
}

//各slistを空リストにした,節点nのgraph型のメモリを確保する関数(Eは0カウント)
graph graph_new(int n){
    graph G;
    NEW(G,1);
    
    G->n=n;
    
    NEW(G->E,G->n);
    int i;
    for(i=0;i<G->n;i++){
        G->E[i]=slist_new();
    }
    
    return G;
}

//リストLの末尾に要素rを追加する関数
void slist_append_tail(slist L,slobj r){
    if(L->head==NULL){
        L->head=r;
        L->tail=r;
    }
    else{
        L->tail->next=r;
        L->tail=r;
    }
    
    return;
}

//節点i(0~n-1)から節点j(0~n-1)に枝を追加する関数
void add_edge(graph G, int i, int j){
    //枝データを追加
    slobj p=slobj_new(j);
    slist_append_tail(G->E[i],p);

    return;
}

//根(節点i)からの長さをdfs
void dfs(graph G, int i, long* d, int* color){
    color[i]=GRAY;
	slobj p=G->E[i]->head;
	while(p!=NULL){
		if(color[p->v]==WHITE){ //未探索なら
			d[p->v]+=d[i];
			dfs(G,p->v,d,color);
		}
		p=p->next;
	}
    color[i]=BLACK;
	return;
}

int main(void){
    int N,Q;
    scanf("%d%d",&N,&Q);

    graph G = graph_new(N);

    //無向グラフに枝を追加
    for(int i=0;i<N-1;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        add_edge(G,a-1,b-1);
        add_edge(G,b-1,a-1);
    }

    //頂点に付いているカウンターの値
    long d[N];
    for(int i=0;i<N;i++) d[i]=0;

    //カウンターの値を標準入力から入れる
    for(int i=0;i<Q;i++){
        int p;
        long x;
        scanf("%d%ld",&p,&x);
        d[p-1]+=x;
    }

    //頂点を訪れたかどうかを格納する配列
    int color[N];
    for(int i=0;i<N;i++) color[i]=0;

    dfs(G,0,d,color);

    for(int i=0;i<N;i++){
        printf("%ld ",d[i]);
    }
    printf("\n");

    return 0;
}