#include<iostream>
#include<cstdio>
#include<cstring>
#define x first
#define y second

template<typename T>inline void read(T &x)
{
	char c=x=0;
	for(c=getchar();!isdigit(c);c=getchar());
	for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48);
}

namespace gg
{
	template<typename T>inline bool check_max(T a,T &b){return a>b?b=a,1:0;}
	template<typename T>inline bool check_min(T a,T &b){return a<b?b=a,1:0;}
	typedef long long ll;
	typedef std::pair<int,int> pii;
	
	inline void insert(int a,pii &b){check_max(a,b.y);if(b.y>b.x)std::swap(b.x,b.y);}

	const int N=201000,M=N*2;
	int begin[N],next[M],to[M];
	int fa[N],dep[N],cnt[N];
	bool ex[N];
	int n,e;

	void add(int x,int y,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		if(k)add(y,x,0);
	}
	pii f[N];
	int df[N];
	void dfs0(int p,int h=0)
	{
		fa[p]=h;
		f[p]=pii(0,0);
		cnt[p]=ex[p];
		dep[p]=0;

		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=h)
			{
				dfs0(q,p);
				insert(f[q].x+1,f[p]);
				check_max(dep[q]+1,dep[p]);
				cnt[p]+=cnt[q];
			}
	}
	void dfs1(int p,int dis=0)
	{
		insert(dis,f[p]);
		df[p]=n+1;
		
		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=fa[p])
			{
				if(cnt[q])check_min(dep[q]+1,df[p]);

				int x=std::max(dis,(f[q].x+1==f[p].x?f[p].y:f[p].x))+1;
				dfs1(q,x);
			}
			else
			{
				if(cnt[p]!=cnt[1])check_min(dis,df[p]);
			}
		if(ex[p])df[p]=0;
	}
	void initialize()
	{
		static char tmp[N];
		read(n);
		for(int i=1,u,v;i<n;i++)
			read(u),read(v),add(u,v);
		scanf("%s",tmp+1);
		for(int i=1;i<=n;i++)ex[i]=tmp[i]=='1';
	}
	void solve()
	{
		initialize();
		dfs0(1),dfs1(1);

//		for(int i=1;i<=n;i++)
//			printf("%d : %d , %d\n",i,f[i].x,f[i].y);

		ll ans=0;
		for(int i=1;i<=n;i++)
		{
			ans+=std::max(0,std::min(f[i].x-1,f[i].y+1)-df[i]+1);
		}
		printf("%lld\n",ans+1);
	}
}
int main()
{
	gg::solve();
	return 0;
}

### Prompt

In CPP, your task is to solve 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<iostream>
#include<cstdio>
#include<cstring>
#define x first
#define y second

template<typename T>inline void read(T &x)
{
	char c=x=0;
	for(c=getchar();!isdigit(c);c=getchar());
	for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48);
}

namespace gg
{
	template<typename T>inline bool check_max(T a,T &b){return a>b?b=a,1:0;}
	template<typename T>inline bool check_min(T a,T &b){return a<b?b=a,1:0;}
	typedef long long ll;
	typedef std::pair<int,int> pii;
	
	inline void insert(int a,pii &b){check_max(a,b.y);if(b.y>b.x)std::swap(b.x,b.y);}

	const int N=201000,M=N*2;
	int begin[N],next[M],to[M];
	int fa[N],dep[N],cnt[N];
	bool ex[N];
	int n,e;

	void add(int x,int y,bool k=1)
	{
		to[++e]=y;
		next[e]=begin[x];
		begin[x]=e;
		if(k)add(y,x,0);
	}
	pii f[N];
	int df[N];
	void dfs0(int p,int h=0)
	{
		fa[p]=h;
		f[p]=pii(0,0);
		cnt[p]=ex[p];
		dep[p]=0;

		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=h)
			{
				dfs0(q,p);
				insert(f[q].x+1,f[p]);
				check_max(dep[q]+1,dep[p]);
				cnt[p]+=cnt[q];
			}
	}
	void dfs1(int p,int dis=0)
	{
		insert(dis,f[p]);
		df[p]=n+1;
		
		for(int i=begin[p],q;i;i=next[i])
			if((q=to[i])!=fa[p])
			{
				if(cnt[q])check_min(dep[q]+1,df[p]);

				int x=std::max(dis,(f[q].x+1==f[p].x?f[p].y:f[p].x))+1;
				dfs1(q,x);
			}
			else
			{
				if(cnt[p]!=cnt[1])check_min(dis,df[p]);
			}
		if(ex[p])df[p]=0;
	}
	void initialize()
	{
		static char tmp[N];
		read(n);
		for(int i=1,u,v;i<n;i++)
			read(u),read(v),add(u,v);
		scanf("%s",tmp+1);
		for(int i=1;i<=n;i++)ex[i]=tmp[i]=='1';
	}
	void solve()
	{
		initialize();
		dfs0(1),dfs1(1);

//		for(int i=1;i<=n;i++)
//			printf("%d : %d , %d\n",i,f[i].x,f[i].y);

		ll ans=0;
		for(int i=1;i<=n;i++)
		{
			ans+=std::max(0,std::min(f[i].x-1,f[i].y+1)-df[i]+1);
		}
		printf("%lld\n",ans+1);
	}
}
int main()
{
	gg::solve();
	return 0;
}

```