#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
typedef long long LL;
const int N=200005;
int n,m1[N],m2[N],sz[N],low[N];char s[N];LL ans;
int pp,lnk[N],nxt[N*2],to[N*2];
void ae(int k1,int k2){to[++pp]=k2,nxt[pp]=lnk[k1],lnk[k1]=pp;}
void dfs(int k1,int k2){
	sz[k1]=s[k1]&15;
	if(s[k1]=='0')low[k1]=0X3F3F3F3F;
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2){
		dfs(to[i],k1);
		int len=m1[to[i]]+1; 
		if(len>m1[k1])m2[k1]=m1[k1],m1[k1]=len;
		else m2[k1]=max(m2[k1],len);
		sz[k1]+=sz[to[i]];
		if(sz[to[i]])low[k1]=min(low[k1],m1[to[i]]+1);
	}
}
void dfs2(int k1,int k2){
	ans+=max(0,min(m1[k1]-1,m2[k1]+1)-low[k1]+1);
//	fprintf(stderr,"%d %d %d %d %lld\n",k1,m1[k1],m2[k1],low[k1],ans);
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2){
		int len=m1[k1]==m1[to[i]]+1?m2[k1]+1:m1[k1]+1;
		if(len>m1[to[i]])m2[to[i]]=m1[to[i]],m1[to[i]]=len;
		else m2[to[i]]=max(m2[to[i]],len);
		if(sz[to[i]]<sz[1])low[to[i]]=min(low[to[i]],len);
		dfs2(to[i],k1);
	}
}
int main(){
	scanf("%d",&n);
	rep(i,2,n){
		int k1,k2;scanf("%d%d",&k1,&k2);
		ae(k1,k2),ae(k2,k1);
	}
	scanf("%s",s+1);
	dfs(1,0),dfs2(1,0);
	printf("%lld\n",ans+1);
	return 0;
}
### Prompt

Generate a Cpp solution to the following problem:

There is a tree with N vertices. The vertices are numbered 1 through N. For each 1 ≤ i ≤ N - 1, the i-th edge connects vertices a_i and b_i. The lengths of all the edges are 1.

Snuke likes some of the vertices. The information on his favorite vertices are given to you as a string s of length N. For each 1 ≤ i ≤ N, s_i is `1` if Snuke likes vertex i, and `0` if he does not like vertex i.

Initially, all the vertices are white. Snuke will perform the following operation exactly once:

* Select a vertex v that he likes, and a non-negative integer d. Then, paint all the vertices black whose distances from v are at most d.



Find the number of the possible combinations of colors of the vertices after the operation.

Constraints

* 2 ≤ N ≤ 2×10^5
* 1 ≤ a_i, b_i ≤ N
* The given graph is a tree.
* |s| = N
* s consists of `0` and `1`.
* s contains at least one occurrence of `1`.

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}
s


Output

Print the number of the possible combinations of colors of the vertices after the operation.

Examples

Input

4
1 2
1 3
1 4
1100


Output

4


Input

5
1 2
1 3
1 4
4 5
11111


Output

11


Input

6
1 2
1 3
1 4
2 5
2 6
100011


Output

8

### Response

```cpp
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
typedef long long LL;
const int N=200005;
int n,m1[N],m2[N],sz[N],low[N];char s[N];LL ans;
int pp,lnk[N],nxt[N*2],to[N*2];
void ae(int k1,int k2){to[++pp]=k2,nxt[pp]=lnk[k1],lnk[k1]=pp;}
void dfs(int k1,int k2){
	sz[k1]=s[k1]&15;
	if(s[k1]=='0')low[k1]=0X3F3F3F3F;
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2){
		dfs(to[i],k1);
		int len=m1[to[i]]+1; 
		if(len>m1[k1])m2[k1]=m1[k1],m1[k1]=len;
		else m2[k1]=max(m2[k1],len);
		sz[k1]+=sz[to[i]];
		if(sz[to[i]])low[k1]=min(low[k1],m1[to[i]]+1);
	}
}
void dfs2(int k1,int k2){
	ans+=max(0,min(m1[k1]-1,m2[k1]+1)-low[k1]+1);
//	fprintf(stderr,"%d %d %d %d %lld\n",k1,m1[k1],m2[k1],low[k1],ans);
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2){
		int len=m1[k1]==m1[to[i]]+1?m2[k1]+1:m1[k1]+1;
		if(len>m1[to[i]])m2[to[i]]=m1[to[i]],m1[to[i]]=len;
		else m2[to[i]]=max(m2[to[i]],len);
		if(sz[to[i]]<sz[1])low[to[i]]=min(low[to[i]],len);
		dfs2(to[i],k1);
	}
}
int main(){
	scanf("%d",&n);
	rep(i,2,n){
		int k1,k2;scanf("%d%d",&k1,&k2);
		ae(k1,k2),ae(k2,k1);
	}
	scanf("%s",s+1);
	dfs(1,0),dfs2(1,0);
	printf("%lld\n",ans+1);
	return 0;
}
```