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;
int main() {
  int n;
  cin >> n;
  int start[n];
  int to[n - 1];
  int next[n - 1];
  memset(start, -1, n * sizeof(int));
  for (int i = 0; i <= n - 2; i++) {
    int p;
    cin >> p;
    to[i] = i + 1;
    next[i] = start[p];
    start[p] = i;
  }
  bool isBlack[n];
  for (int i = 0; i <= n; i++) {
    cin >> isBlack[i];
  }
  long long dp[n][2];
  memset(dp, 0, sizeof(long long) * n * 2);
  for (int i = n - 1; i >= 0; i--) {
    dp[i][isBlack[i]] = 1;
    for (int e = start[i]; e != -1; e = next[e]) {
      int child = to[e];
      dp[i][1] =
          dp[i][1] * (dp[child][0] + dp[child][1]) + dp[i][0] * dp[child][1];
      dp[i][1] %= 1000 * 1000 * 1000 + 7;
      dp[i][0] = dp[i][0] * (dp[child][0] + dp[child][1]);
      dp[i][0] %= 1000 * 1000 * 1000 + 7;
    }
  }
  cout << dp[0][1] << endl;
  return 0;
}
