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;
int mod = (int)1e9 + 7;
void dfs(int v, int p, list<int>* adj, int* c, long long** dp) {
  if (c[v] == 1) {
    dp[0][v] = 0;
    dp[1][v] = 1;
  } else {
    dp[0][v] = 1;
    dp[1][v] = 0;
  }
  for (auto i : adj[v]) {
    if (i == p) continue;
    dfs(i, v, adj, c, dp);
    long long zero = dp[0][v], one = dp[1][v];
    dp[0][v] = zero * dp[0][i];
    dp[0][v] %= mod;
    dp[1][v] = (one * dp[0][i]) % mod + (zero * dp[1][i]) % mod;
    dp[1][v] %= mod;
    dp[0][v] += zero * dp[1][i];
    dp[0][v] %= mod;
    dp[1][v] += one * dp[1][i];
    dp[1][v] %= mod;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  list<int>* adj = new list<int>[n];
  for (int i = 1; i < n; i++) {
    int a;
    cin >> a;
    adj[i].push_back(a);
    adj[a].push_back(i);
  }
  int* c = new int[n];
  for (int i = 0; i < n; i++) {
    cin >> c[i];
  }
  long long** dp = new long long*[2];
  dp[0] = new long long[n]();
  dp[1] = new long long[n]();
  dfs(0, -1, adj, c, dp);
  cout << dp[1][0] << endl;
  return 0;
}
