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<vector<int>> graph;
vector<int> isBlack;
vector<long long> dp[2];
long long previous[2];
void dfs(int curr) {
  dp[0][curr] = 1 - isBlack[curr];
  dp[1][curr] = isBlack[curr];
  for (int child : graph[curr]) {
    dfs(child);
    previous[0] = dp[0][curr];
    previous[1] = dp[1][curr];
    dp[0][curr] = 0;
    dp[1][curr] = 0;
    dp[1][curr] =
        (previous[1] * (dp[0][child] + dp[1][child]) % 1000000007) % 1000000007;
    if (isBlack[curr] == 0) {
      dp[0][curr] = (previous[0] * (dp[0][child] + dp[1][child]) % 1000000007) %
                    1000000007;
      dp[1][curr] = (dp[1][curr] + (previous[0] * dp[1][child]) % 1000000007) %
                    1000000007;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  graph.resize(n);
  isBlack.resize(n);
  dp[1].resize(n);
  dp[0].resize(n);
  for (int i = 0; i < n - 1; i++) {
    int p;
    cin >> p;
    graph[p].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    cin >> isBlack[i];
  }
  dfs(0);
  cout << dp[1][0] << '\n';
  return 0;
}
