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 long long MOD = 1000000007;
int n;
long long dp[100005][2];
int c[100005];
vector<int> edge[100005];
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x;
    cin >> x;
    edge[x].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = n - 1; i >= 0; i--) {
    long long a0 = 0, a1 = 0;
    if (c[i] == 0)
      a0++;
    else
      a1++;
    for (int j = 0; j < edge[i].size(); j++) {
      int x = edge[i][j];
      long long a01 = (a0 * dp[x][0] + a0 * dp[x][1]) % MOD;
      long long a11 = (a1 * dp[x][0] + a1 * dp[x][1] + a0 * dp[x][1]) % MOD;
      a0 = a01;
      a1 = a11;
    }
    dp[i][0] = a0;
    dp[i][1] = a1;
  }
  cout << dp[0][1];
  return 0;
}
