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;
void EE(int a, int b, int& x, int& y) {
  if (a % b == 0) {
    x = 0;
    y = 1;
    return;
  }
  EE(b, a % b, x, y);
  int temp = x;
  x = y;
  y = temp - y * (a / b);
}
int inverse(int a, int m) {
  int x, y;
  EE(a, m, x, y);
  if (x < 0) x += m;
  return x;
}
int n;
vector<int> g[100005];
int dp[100005][2];
vector<int> x;
void dfs(int u, int par) {
  dp[u][0] = 1;
  if (x[u] == 1) {
    dp[u][0] = 0;
    dp[u][1] = 1;
  }
  for (auto v : g[u]) {
    if (v != par) {
      dfs(v, u);
      if (x[u] == 1) {
        dp[u][1] = (1LL * dp[u][1] * (dp[v][0] + dp[v][1])) % mod;
      } else {
        dp[u][0] = (1LL * dp[u][0] * (dp[v][0] + dp[v][1])) % mod;
      }
    }
  }
  if (x[u] != 1) {
    int val = dp[u][0];
    dp[u][1] = 0;
    for (auto v : g[u]) {
      if (v != par) {
        int temp = (1LL * val * inverse(dp[v][0] + dp[v][1], mod)) % mod;
        temp = (1LL * temp * dp[v][1]) % mod;
        dp[u][1] = (dp[u][1] + temp) % mod;
      }
    }
  }
  return;
}
int main() {
  cin >> n;
  x.resize(n);
  for (int i = 0, v; i < n - 1; i++) {
    cin >> v;
    g[i + 1].push_back(v);
    g[v].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    cin >> x[i];
  }
  dfs(0, -1);
  cout << dp[0][1] << endl;
  return 0;
}
