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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

#define A first
#define B second

const int MAXN = 2e5+5;

ll ans = 0;

int n;
vector<int> g[MAXN];
pii a[MAXN][2];
string s;

void dfsdown(int v, int f = -1) {
	a[v][0] = pii(0, v);
	for (int u : g[v]) {
		if (u == f)
			continue;
		dfsdown(u, v);
		pii t = a[u][0];
		t.A++;
		if (t > a[v][0]) {
			a[v][1] = a[v][0];
			a[v][0] = t;
		} else if (t > a[v][1])
			a[v][1] = t;
	}
}

void dfsup(int v, int f = -1) {
	for (int u : g[v]) {
		if (u == f)
			continue;
		pii t;
		if (a[v][0].B == u)
			t = a[v][1];
		else
			t = a[v][0];
		t.A++;

		if (t > a[u][0]) {
			a[u][1] = a[u][0];
			a[u][0] = t;
		} else
			a[u][0] = t;

		dfsup(u, v);
	}
}

int main() {
	cin >> n;
	for (int i = 0; i < n-1; ++i) {
		int a, b;
		cin >> a >> b;
		a--; b--;

		g[a].push_back(b);
		g[b].push_back(a);
	}

	cin >> s;

	dfsdown(0, -1);
	dfsup(0, -1);
	
	for (int i = 0; i < n; ++i)
		ans += a[i][1].A + 1;

	/* for (int i = 0; i < n; ++i) {
		for (int u : g[i]) {
			if (u > i)
				continue;
			int vu, vi;
			if (i == a[u][0].B)
				vu = a[u][1].A;
			else
				vu = a[u][0].A;

			if (u == a[i][0].B)
				vi = a[i][1].A;
			else
				vi = a[i][0].A;
		}
	} */

	cout << ans + (n-1) << endl;
}