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[100010];
int arr[100010] = {0};
int vis[100010] = {0};
long long dp[100010][2];
void dfs(int u) {
  vis[u] = 1;
  dp[u][0] = 1 - arr[u];
  dp[u][1] = arr[u];
  vector<int>::iterator vec;
  for (vec = graph[u].begin(); vec != graph[u].end(); vec++) {
    int i = (*vec);
    if (vis[i])
      continue;
    else {
      dfs(i);
      long long zero, one;
      zero = dp[u][0];
      one = dp[u][1];
      dp[u][0] = (zero * dp[i][1]) % 1000000007;
      dp[u][1] = (one * dp[i][1]) % 1000000007;
      dp[u][0] += (zero * dp[i][0]) % 1000000007;
      dp[u][0] = dp[u][0] % 1000000007;
      dp[u][1] += (zero * dp[i][1] + one * dp[i][0]) % 1000000007;
      dp[u][1] = dp[u][1] % 1000000007;
    }
  }
}
int main() {
  int n, v;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> v;
    graph[v].push_back(i);
    graph[i].push_back(v);
  }
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  dfs(0);
  cout << dp[0][1] << endl;
}
