Problem: We have a tree with N vertices. The vertices are numbered 1, 2, ..., N. The i-th (1 ≦ i ≦ N - 1) edge connects the two vertices A_i and B_i.

Takahashi wrote integers into K of the vertices. Specifically, for each 1 ≦ j ≦ K, he wrote the integer P_j into vertex V_j. The remaining vertices are left empty. After that, he got tired and fell asleep.

Then, Aoki appeared. He is trying to surprise Takahashi by writing integers into all empty vertices so that the following condition is satisfied:

* Condition: For any two vertices directly connected by an edge, the integers written into these vertices differ by exactly 1.



Determine if it is possible to write integers into all empty vertices so that the condition is satisfied. If the answer is positive, find one specific way to satisfy the condition.

Constraints

* 1 ≦ N ≦ 10^5
* 1 ≦ K ≦ N
* 1 ≦ A_i, B_i ≦ N (1 ≦ i ≦ N - 1)
* 1 ≦ V_j ≦ N (1 ≦ j ≦ K) (21:18, a mistake in this constraint was corrected)
* 0 ≦ P_j ≦ 10^5 (1 ≦ j ≦ K)
* The given graph is a tree.
* All v_j are distinct.

Input

The input is given from Standard Input in the following format:


N
A_1 B_1
A_2 B_2
:
A_{N-1} B_{N-1}
K
V_1 P_1
V_2 P_2
:
V_K P_K


Output

If it is possible to write integers into all empty vertices so that the condition is satisfied, print `Yes`. Otherwise, print `No`.

If it is possible to satisfy the condition, print N lines in addition. The v-th (1 ≦ v ≦ N) of these N lines should contain the integer that should be written into vertex v. If there are multiple ways to satisfy the condition, any of those is accepted.

Examples

Input

5
1 2
3 1
4 3
3 5
2
2 6
5 7


Output

Yes
5
6
6
5
7


Input

5
1 2
3 1
4 3
3 5
3
2 6
4 3
5 7


Output

No


Input

4
1 2
2 3
3 4
1
1 0


Output

Yes
0
-1
-2
-3

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,flag,tot,hh,dep[500001],a[500001],e[500001],nt[500001],hd[500001],v[500001],l[500001],r[500001];
void build(int x,int y){
	tot++;
	e[tot]=y;
	nt[tot]=hd[x];hd[x]=tot;
}
void dfs(int x,int fa){
	int i;
	dep[x]=dep[fa]+1;
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		dfs(e[i],x);
	}
}
void get(int x,int fa){
	int i;
	if(v[x])l[x]=r[x]=a[x];
	 else l[x]=-1e9,r[x]=1e9;
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		get(e[i],x);
		l[x]=max(l[x],l[e[i]]-1);
		r[x]=min(r[x],r[e[i]]+1);
	}
	if(l[x]>r[x])flag=1;
}
void ha(int x,int fa){
	int i;
	if(!v[x]){
		if(a[fa]+1>=l[x]&&a[fa]+1<=r[x])a[x]=a[fa]+1;
		 else a[x]=a[fa]-1;
	}
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		ha(e[i],x);
	}
}
int main(){
	int i,x,y,m;
	scanf("%d",&n);
	for(i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		build(x,y);build(y,x);
	}
	dfs(1,0);
	scanf("%d",&m);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		a[x]=y;v[x]=1;
		if(i==1)hh=(y&1)^(dep[x]&1);
		 else if(((y&1)^(dep[x]&1))!=hh)return puts("No"),0;
	}
	get(1,0);
	if(flag)return puts("No"),0;
	if(!v[1])v[1]=1,a[1]=l[1];
	ha(1,0);
	puts("Yes");
	for(i=1;i<=n;i++)printf("%d\n",a[i]);
}
/*
考虑不合法的情况就是距离为偶数的点权奇偶不同或为奇数的奇偶相同。
判掉之后每个点的取值范围可从其儿子节点递推过来 
*/