#define rep(i, n) for(ll i = 0; i < n; i++)
#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using P = pair<int, int>;

vector<ll> to[200001];
vector<ll> ans;
void dfs(ll v, ll p=-1) {
	for (ll u : to[v]) {
		if (u == p) continue;
		ans[u] += ans[v];
		dfs(u,v);
	}
}


int main() {
	ll n,q;
	cin >> n >> q;
	rep(i,n-1) {
		ll a,b;
		cin >> a >> b;
		a--,b--;
		to[a].push_back(b);
		to[b].push_back(a);
	}
	ans.resize(n);
	rep(i,q) {
		ll p,x;
		cin >> p >> x;
		p--;
		ans[p] += x;
	}
	dfs(0);

	rep(i,n) {
		cout << ans[i] << endl;
	}

	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
#define rep(i, n) for(ll i = 0; i < n; i++)
#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using P = pair<int, int>;

vector<ll> to[200001];
vector<ll> ans;
void dfs(ll v, ll p=-1) {
	for (ll u : to[v]) {
		if (u == p) continue;
		ans[u] += ans[v];
		dfs(u,v);
	}
}


int main() {
	ll n,q;
	cin >> n >> q;
	rep(i,n-1) {
		ll a,b;
		cin >> a >> b;
		a--,b--;
		to[a].push_back(b);
		to[b].push_back(a);
	}
	ans.resize(n);
	rep(i,q) {
		ll p,x;
		cin >> p >> x;
		p--;
		ans[p] += x;
	}
	dfs(0);

	rep(i,n) {
		cout << ans[i] << endl;
	}

	return 0;

}

```