#include <bits/stdc++.h>
using namespace std;

int main(){
  int N,Q;
  cin >> N >> Q;
  vector<pair<int,int>> A(N-1,pair<int,int>(0,0));
  for(int i=0;i<N-1;i++){
    cin >> A.at(i).first >> A.at(i).second;
  }
  sort(A.begin(),A.end());
  vector<long long> P(N+1,0);
  int r,s;
  for(int i=0;i<Q;i++){
    cin >> r >> s;
    P.at(r) += s;
  }
  r=0;
  s=0;
  for(int i=0;i<N-1;i++){
    r=A.at(i).first;
    s=A.at(i).second;
    P.at(s) += P.at(r);
  }
  for(int i=1;i<=N;i++){
    cout << P.at(i) << " " ;
  }
}
### Prompt

Create a solution in CPP for 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 <bits/stdc++.h>
using namespace std;

int main(){
  int N,Q;
  cin >> N >> Q;
  vector<pair<int,int>> A(N-1,pair<int,int>(0,0));
  for(int i=0;i<N-1;i++){
    cin >> A.at(i).first >> A.at(i).second;
  }
  sort(A.begin(),A.end());
  vector<long long> P(N+1,0);
  int r,s;
  for(int i=0;i<Q;i++){
    cin >> r >> s;
    P.at(r) += s;
  }
  r=0;
  s=0;
  for(int i=0;i<N-1;i++){
    r=A.at(i).first;
    s=A.at(i).second;
    P.at(s) += P.at(r);
  }
  for(int i=1;i<=N;i++){
    cout << P.at(i) << " " ;
  }
}
```