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;
const unsigned long long MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
int n;
int *color;
vector<int> *child;
unsigned long long dp[MAXN][2];
int main() {
  scanf("%d", &n);
  child = new vector<int>[n];
  for (int i = 0; i < n - 1; ++i) {
    int x;
    scanf("%d", &x);
    child[x].push_back(i + 1);
  }
  color = new int[n];
  for (int i = 0; i < n; ++i) {
    scanf("%d", color + i);
  }
  for (int i = n - 1; i >= 0; --i) {
    if (color[i] == 0) {
      dp[i][0] = 1;
      dp[i][1] = 0;
      for (int k = 0; k < child[i].size(); ++k) {
        int u = child[i][k];
        dp[i][1] =
            (dp[i][0] * dp[u][1] + dp[i][1] * (dp[u][0] + dp[u][1])) % MOD;
        dp[i][0] = (dp[i][0] * (dp[u][0] + dp[u][1])) % MOD;
      }
    } else {
      dp[i][0] = 0;
      dp[i][1] = 1;
      for (int k = 0; k < child[i].size(); ++k) {
        int u = child[i][k];
        dp[i][1] = (dp[i][1] * (dp[u][0] + dp[u][1])) % MOD;
      }
    }
  }
  cout << dp[0][1] << endl;
  delete[] color;
  delete[] child;
  return 0;
}
