#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 200005;
int n, a, b, q, p[MX], ans[MX];
vector<int> G[MX];
bool used[MX];
void dfs(int v, int x){
	used[v] = true;
	x += p[v];
	ans[v] = x;
	for(int t: G[v]) if(!used[t]) dfs(t, x);
}

int main(){
	scanf("%d%d", &n, &q);
	rep(i,n-1){
		scanf("%d%d", &a, &b);a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	rep(i,q){
		scanf("%d%d", &a, &b); a--;
		p[a] += b;
	}
	dfs(0,0);
	rep(i,n) printf("%d%c", ans[i], i == n-1 ? '\n' : ' ');
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

Examples

Input

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


Output

100 110 111 110


Input

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


Output

20 20 20 20 20 20

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 200005;
int n, a, b, q, p[MX], ans[MX];
vector<int> G[MX];
bool used[MX];
void dfs(int v, int x){
	used[v] = true;
	x += p[v];
	ans[v] = x;
	for(int t: G[v]) if(!used[t]) dfs(t, x);
}

int main(){
	scanf("%d%d", &n, &q);
	rep(i,n-1){
		scanf("%d%d", &a, &b);a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	rep(i,q){
		scanf("%d%d", &a, &b); a--;
		p[a] += b;
	}
	dfs(0,0);
	rep(i,n) printf("%d%c", ans[i], i == n-1 ? '\n' : ' ');
}
```