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 int MOD = 1e9 + 7, N_MAX = 1e5;
int n;
vector<int> gr[N_MAX + 2];
bool col[N_MAX + 2];
int dp[N_MAX + 2][2];
void dfs(int arg) {
  dp[arg][0] = !col[arg];
  dp[arg][1] = col[arg];
  for (auto it : gr[arg]) {
    dfs(it);
    int aux0 = dp[arg][0], aux1 = dp[arg][1];
    dp[arg][0] = (1LL * aux0 * dp[it][0] + 1LL * aux0 * dp[it][1]) % MOD;
    dp[arg][1] = (1LL * aux0 * dp[it][1] + 1LL * aux1 * dp[it][0] +
                  1LL * aux1 * dp[it][1]) %
                 MOD;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x;
    cin >> x;
    gr[x].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> col[i];
  dfs(0);
  cout << dp[0][1];
  return 0;
}
