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

ll n,x,y,q;
vector <ll> adj[200009];
ll an[200009];
ll vl[200009];

void btr(ll x,ll s,ll pr){
	an[x]=vl[x]+s;
	
	for(int i=0;i<adj[x].size();i++)
	if(adj[x][i]!=pr)
	btr(adj[x][i],s+vl[x],x);
}

int main(){
	cin>>n>>q;
	
	for(int i=1;i<n;i++)
	cin>>x>>y,adj[x].push_back(y),adj[y].push_back(x);
	
	for(int i=0;i<q;i++)
	cin>>x>>y,vl[x]+=y;
	
	btr(1,0,-1);
	
	for(int i=1;i<=n;i++)
	cout<<an[i]<<" ";
}
### Prompt

Please create a solution in Cpp 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
# define ll long long
# include <bits/stdc++.h>
using namespace std;

ll n,x,y,q;
vector <ll> adj[200009];
ll an[200009];
ll vl[200009];

void btr(ll x,ll s,ll pr){
	an[x]=vl[x]+s;
	
	for(int i=0;i<adj[x].size();i++)
	if(adj[x][i]!=pr)
	btr(adj[x][i],s+vl[x],x);
}

int main(){
	cin>>n>>q;
	
	for(int i=1;i<n;i++)
	cin>>x>>y,adj[x].push_back(y),adj[y].push_back(x);
	
	for(int i=0;i<q;i++)
	cin>>x>>y,vl[x]+=y;
	
	btr(1,0,-1);
	
	for(int i=1;i<=n;i++)
	cout<<an[i]<<" ";
}
```