Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices are colored white.

Consider a set consisting of k (0 ≤ k < n) edges of Appleman's tree. If Appleman deletes these edges from the tree, then it will split into (k + 1) parts. Note, that each part will be a tree with colored vertices.

Now Appleman wonders, what is the number of sets splitting the tree in such a way that each resulting part will have exactly one black vertex? Find this number modulo 1000000007 (109 + 7).

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of tree vertices. 

The second line contains the description of the tree: n - 1 integers p0, p1, ..., pn - 2 (0 ≤ pi ≤ i). Where pi means that there is an edge connecting vertex (i + 1) of the tree and vertex pi. Consider tree vertices are numbered from 0 to n - 1.

The third line contains the description of the colors of the vertices: n integers x0, x1, ..., xn - 1 (xi is either 0 or 1). If xi is equal to 1, vertex i is colored black. Otherwise, vertex i is colored white.

Output

Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Examples

Input

3
0 0
0 1 1


Output

2


Input

6
0 1 1 0 4
1 1 0 0 1 0


Output

1


Input

10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1


Output

27
#include <bits/stdc++.h>
using namespace std;
queue<int> tree[100000];
int colour[100000];
long long int DP[2][100000];
void generateAnswer(int root) {
  if (tree[root].size() == 0) {
    if (colour[root] == 1) {
      DP[1][root] = DP[0][root];
    }
    return;
  }
  while (!tree[root].empty()) {
    int curr = tree[root].front();
    generateAnswer(curr);
    tree[root].pop();
    DP[1][root] = (DP[1][root] * DP[0][curr]) % 1000000007;
    DP[1][root] += (DP[0][root] * DP[1][curr]) % 1000000007;
    DP[0][root] = (DP[0][root] * DP[0][curr]) % 1000000007;
  }
  if (colour[root] == 1) {
    DP[1][root] = DP[0][root];
  } else {
    DP[0][root] = (DP[0][root] + DP[1][root]) % 1000000007;
  }
  return;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < 100000; i++) {
    colour[i] = -1;
    DP[0][i] = 1;
    DP[1][i] = 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int from;
    cin >> from;
    tree[from].push(i + 1);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &colour[i]);
  }
  generateAnswer(0);
  printf("%lld", DP[1][0] % 1000000007);
  return 0;
}
