Problem:
Score : 1900 points

Problem Statement

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.

Partial Score

In the test set worth 1300 points, s consists only 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.

Sample Input 1

4
1 2
1 3
1 4
1100

Sample Output 1

4

The following four combinations of colors of the vertices are possible:

Sample Input 2

5
1 2
1 3
1 4
4 5
11111

Sample Output 2

11

This case satisfies the additional constraint for the partial score.

Sample Input 3

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

Sample Output 3

8

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

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

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N;
vint G[222222];
string S;
void dfs(int v,int p,int d,vint &dist){
    dist[v]=d;
    for(auto u:G[v]){
        if(u==p)continue;
        dfs(u,v,d+1,dist);
    }
}


    /*
    vint lim(N);
    rep(i,N)lim[i]=max(distS[i],distT[i])-1;
    rep(i,N)chmin(lim[i],R[i]+1);
*/


int ans;
int R[222222];
int dw[222222];


vint distS,distT;
void dfs2(int v,int p){
    R[v]=0;
    dw[v]=INT_MAX;
    for(auto u:G[v]){
        if(u==p)continue;
        dfs2(u,v);
        chmax(R[v],R[u]+1);
    }
    if(S[v]=='1')dw[v]=0;

    int lim=max(distS[v],distT[v])-1;
    chmin(lim,R[v]+1);
    ans+=max(0ll,lim-dw[v]+1);
    if(p!=-1)chmin(dw[p],max(dw[v]-1,R[v]+1));
}



signed main(){
    cin>>N;
    rep(i,N-1){
        int a,b;
        cin>>a>>b;
        a--;b--;G[a].pb(b);G[b].pb(a);
    }
    cin>>S;

    distS.resize(N);distT.resize(N);
    dfs(0,-1,0,distS);
    int s=0;
    rep(i,N)if(distS[i]>distS[s])s=i;
    dfs(s,-1,0,distS);
    int t=0;
    rep(i,N)if(distS[i]>distS[t])t=i;
    dfs(t,-1,0,distT);

    vint lis;
    rep(i,N){
        if(max(distS[i],distT[i])==(distS[t]+1)/2)lis.pb(i);
    }

    if(lis.size()==1){
        dfs2(lis[0],-1);
    }
    else{
        dfs2(lis[0],lis[1]);
        dfs2(lis[1],lis[0]);
    }
    cout<<ans+1<<endl;


    return 0;
}