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;
using namespace std;
long long n, mark[200009], dp[200009][2], col[200009], mod = 1e9 + 7, ans = 1,
                                                       x;
vector<long long> g[200009];
long long xgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1, gcd = xgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - (long long)(a / b) * y1;
  return gcd;
}
long long dfs(long long x1) {
  if (col[x1] == 1) {
    dp[x1][0] = 0;
    dp[x1][1] = 1;
    return 0;
  }
  mark[x1] = 1;
  vector<long long> v1, v2;
  long long pr = 1;
  for (long long i = 0; i < g[x1].size(); i++) {
    if (mark[g[x1][i]] == 0) {
      dfs(g[x1][i]);
      pr *= ((dp[g[x1][i]][0] % mod) + (dp[g[x1][i]][1] % mod)) % mod;
      pr %= mod;
      v1.push_back((dp[g[x1][i]][0] + dp[g[x1][i]][1]) % mod);
      v2.push_back(dp[g[x1][i]][1] % mod);
    }
  }
  dp[x1][0] = pr % mod;
  for (long long i = 0; i < v1.size(); i++) {
    long long a, x = 1, y = 1, g1 = xgcd(v1[i], mod, x, y);
    a = (dp[x1][0] * ((x + mod) % mod)) % mod;
    dp[x1][1] += (a * v2[i]) % mod;
    dp[x1][1] %= mod;
  }
  return 0;
}
int main() {
  cin >> n;
  for (long long i = 1; i < n; i++) {
    cin >> x;
    g[x].push_back(i);
    g[i].push_back(x);
  }
  for (long long i = 0; i < n; i++) cin >> col[i];
  for (long long i = 0; i < n; i++) {
    if (mark[i] == 0 && col[i] == 0) {
      dfs(i);
      ans *= dp[i][1];
      ans %= mod;
    }
  }
  cout << ans;
}
