#include<iostream>
#include<cstdio>
#include<vector>
const int N=2e5+5;
std::vector<int> g[N];
int ans[N],n,q;
void dfs(int u,int fa,int x) {
    ans[u]+=x;
    for(int v:g[u]) if(v!=fa) dfs(v,u,ans[u]);
}
int main() {
    scanf("%d%d",&n,&q);
    for(int i=1,x,y;i<n;i++) {
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    for(int x,y;q--;) {
        scanf("%d%d",&x,&y);
        ans[x]+=y;
    }
    dfs(1,1,0);
    for(int i=1;i<=n;i++) printf("%d%c",ans[i]," \n"[i==n]);
    return 0;
}
### Prompt

Please formulate a CPP solution to the following problem:

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<iostream>
#include<cstdio>
#include<vector>
const int N=2e5+5;
std::vector<int> g[N];
int ans[N],n,q;
void dfs(int u,int fa,int x) {
    ans[u]+=x;
    for(int v:g[u]) if(v!=fa) dfs(v,u,ans[u]);
}
int main() {
    scanf("%d%d",&n,&q);
    for(int i=1,x,y;i<n;i++) {
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    for(int x,y;q--;) {
        scanf("%d%d",&x,&y);
        ans[x]+=y;
    }
    dfs(1,1,0);
    for(int i=1;i<=n;i++) printf("%d%c",ans[i]," \n"[i==n]);
    return 0;
}
```