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;
long long int inv(long long int a) {
  int n = 1000000005;
  int p = 1000000007;
  long long int ret = 1;
  while (n != 0) {
    if (n % 2 == 1) ret = (ret * a) % p;
    n /= 2;
    a = a * a % p;
  }
  return ret;
}
int main() {
  int n;
  cin >> n;
  long long int dp[n][2];
  int i, j;
  for (i = 0; i < n; i++) {
    dp[i][0] = 0;
    dp[i][1] = 0;
  }
  vector<int> child[n];
  long long int temp;
  for (i = 1; i < n; i++) {
    cin >> temp;
    child[temp].push_back(i);
  }
  int wb[n];
  for (i = 0; i < n; i++) {
    cin >> wb[i];
  }
  for (i = n - 1; i >= 0; i--) {
    if (child[i].size() == 0) {
      dp[i][wb[i]] = 1;
    } else {
      temp = 1;
      for (j = 0; j < child[i].size(); j++) {
        temp *= ((dp[child[i][j]][0] + dp[child[i][j]][1]) % 1000000007);
        temp %= 1000000007;
      }
      if (wb[i] == 1) {
        dp[i][1] = temp;
        dp[i][0] = 0;
      } else {
        dp[i][0] = temp;
        dp[i][1] = 0;
        for (j = 0; j < child[i].size(); j++) {
          dp[i][1] +=
              temp * dp[child[i][j]][1] % 1000000007 *
              inv((dp[child[i][j]][0] + dp[child[i][j]][1]) % 1000000007) %
              1000000007;
          dp[i][1] %= 1000000007;
        }
      }
    }
  }
  cout << dp[0][1] << "\n";
  return 0;
}
