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 M = 100005;
const int mod = 1e9 + 7;
vector<int> adj[M];
int n, color[M];
long long dp[M][3];
void dfs(int node, int p) {
  dp[node][1] = color[node];
  dp[node][0] = (color[node] == 0);
  for (auto i : adj[node]) {
    if (i == p) continue;
    dfs(i, node);
    dp[node][1] = dp[node][1] * dp[i][0] + dp[node][0] * dp[i][1] +
                  dp[node][1] * dp[i][1];
    dp[node][1] %= mod;
    dp[node][0] = dp[node][0] * dp[i][0] + dp[node][0] * dp[i][1];
    dp[node][0] %= mod;
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int a;
    cin >> a;
    adj[a + 1].push_back(i);
    adj[i].push_back(a + 1);
  }
  for (int i = 1; i <= n; ++i) cin >> color[i];
  dfs(1, 0);
  cout << dp[1][1] << "\n";
  return 0;
}
