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 <cctype>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long int64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 2e5+5, INF = 0x3f3f3f3f;
struct Edge
{
    int next, to;
    Edge(int next = 0, int to = 0):next(next), to(to) {};
}edge[N<<1];
int n, tot, head[N];
void _add(int x, int y) { edge[++tot] = Edge(head[x], y); head[x] = tot; }
void add(int x, int y) { _add(x, y); _add(y, x); }
int s[N], d1[N], d2[N], d[N];
char c[N];
int64 ans;
void dfs1(int x, int fa)
{
    c[x] == '1'?d[x] = 0, s[x] = 1:d[x] = INF;
    for(int i = head[x]; i; i = edge[i].next)
    {
        int y = edge[i].to; if(y == fa) continue;
        dfs1(y, x), s[x] += s[y];
        if(d1[y]+1 > d1[x]) d2[x] = d1[x], d1[x] = d1[y]+1;
        else if(d1[y]+1 > d2[x]) d2[x] = d1[y]+1;
        if(s[y]) d[x] = min(d[x], d1[y]+1);
    }
}
void dfs2(int x, int fa)
{
    int r = min(d2[x]+1, d1[x]-1);
    if(d[x] <= r) ans += r-d[x]+1;
    for(int i = head[x]; i; i = edge[i].next)
    {
        int y = edge[i].to; if(y == fa) continue;
        int k = d1[y]+1 == d1[x]?d2[x]+1:d1[x]+1;
        if(k > d1[y]) d2[y] = d1[y], d1[y] = k;
        else if(k > d2[y]) d2[y] = k;
        if(s[1]-s[y]) d[y] = min(d[y], k);
        dfs2(y, x);      
    }
}
int main()
{
    n = read();
    for(int i = 1, x, y; i < n; ++i)
        x = read(), y = read(), add(x, y);
    scanf("%s", c+1), dfs1(1, 0), dfs2(1, 0);
    printf("%lld\n", ans+1);
    return 0;
}