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> graph[100005];
bool visited[100005];
bool black[100005];
long long int dp[100005][2];
long long int p = 1000000007;
void dfs(int n) {
  visited[n] = 1;
  if (black[n]) {
    dp[n][1] = 1;
    dp[n][0] = 0;
  } else {
    dp[n][0] = 1;
    dp[n][1] = 0;
  }
  for (int i = 0; i < graph[n].size(); i++) {
    if (visited[graph[n][i]]) {
      continue;
    } else {
      dfs(graph[n][i]);
      if (black[n] == 1) {
        dp[n][1] = (dp[n][1] * (dp[graph[n][i]][0] + dp[graph[n][i]][1])) % p;
      } else {
        dp[n][1] = (dp[n][1] * (dp[graph[n][i]][0] + dp[graph[n][i]][1])) % p;
        dp[n][1] += (dp[graph[n][i]][1] * dp[n][0]) % p;
        dp[n][0] = (dp[n][0] * (dp[graph[n][i]][0] + dp[graph[n][i]][1])) % p;
      }
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int z;
    cin >> z;
    graph[i + 1].push_back(z);
    graph[z].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    cin >> black[i];
  }
  dfs(0);
  cout << ((dp[0][1]) % p);
}
