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> adj[200001];
vector<int> x(200001);
vector<vector<long long> > dp(200001, vector<long long>(2));
void dfs(int node, int parent) {
  int i;
  dp[node][0] = 1;
  dp[node][1] = 0;
  for (i = 0; i < adj[node].size(); i++) {
    if (adj[node][i] == parent) continue;
    dfs(adj[node][i], node);
    dp[node][1] = (dp[node][1] * dp[adj[node][i]][0]) % 1000000007;
    dp[node][1] =
        (dp[node][1] + (dp[node][0] * dp[adj[node][i]][1]) % 1000000007) %
        1000000007;
    dp[node][0] = (dp[node][0] * dp[adj[node][i]][0]) % 1000000007;
  }
  if (x[node] == 1)
    dp[node][1] = dp[node][0];
  else
    dp[node][0] = (dp[node][0] + dp[node][1]) % 1000000007;
}
int main() {
  long long n, i, u, v;
  cin >> n;
  for (i = 1; i <= n - 1; i++) {
    cin >> u;
    adj[i].push_back(u);
    adj[u].push_back(i);
  }
  for (i = 0; i < n; i++) {
    cin >> u;
    x[i] = u;
  }
  dfs(0, -1);
  cout << dp[0][1] << " ";
  return 0;
}
