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;
vector<int> adj[100010];
long long int col[100010];
long long int dp[100010][2];
void dfs(int x, int f) {
  if (col[x] == 1) {
    dp[x][0] = 0;
    dp[x][1] = 1;
  } else {
    dp[x][0] = 1;
    dp[x][1] = 0;
  }
  for (int i = 0; i < adj[x].size(); i++) {
    int y = adj[x][i];
    if (y != f) {
      dfs(y, x);
      long long int no_black = dp[x][0];
      long long int one_black = dp[x][1];
      dp[x][0] = dp[x][1] = 0;
      if (dp[y][1] > 0) {
        dp[x][1] += one_black * dp[y][1];
        dp[x][1] += no_black * dp[y][1];
        dp[x][0] += no_black * dp[y][1];
      }
      if (dp[y][0] > 0) {
        dp[x][1] += one_black * dp[y][0];
        dp[x][0] += no_black * dp[y][0];
      }
      dp[x][0] = dp[x][0] % 1000000007;
      dp[x][1] = dp[x][1] % 1000000007;
    }
  }
  return;
}
int main() {
  int n, x;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    cin >> x;
    adj[i].push_back(x);
    adj[x].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0, -1);
  cout << dp[0][1] << endl;
  return 0;
}
