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 long long int MOD = 1000000007, MaxN = 100001;
long long int f[MaxN], g[MaxN], C[MaxN], A, B;
vector<int> ady[MaxN];
long long int mcd(long long int a, long long int b) {
  if (!b) {
    A = 1, B = 0;
    return a;
  }
  long long int gcd = mcd(b, a % b), x, y;
  x = A, y = B;
  A = y, B = x - y * (a / b);
  return gcd;
}
long long int inverse(long long int a) {
  mcd(a, MOD);
  return (A + MOD * 5) % MOD;
}
void DFS(int x, int past) {
  long long int s = 1, i, j, p, q, T = 1;
  for (i = 0; i < ady[x].size(); i++) {
    if (ady[x][i] == past) continue;
    p = ady[x][i], DFS(p, x);
    T = (T * (g[p] + f[p]) % MOD) % MOD;
  }
  if (C[x]) {
    f[x] = T;
    g[x] = 0;
  } else {
    for (i = 0; i < ady[x].size(); i++) {
      if (ady[x][i] == past) continue;
      p = ady[x][i];
      f[x] = (f[x] + ((T * f[p]) % MOD) * inverse((f[p] + g[p]) % MOD)) % MOD;
    }
    g[x] = T;
  }
}
int main() {
  long long int i, j, p, q, N;
  cin >> N;
  for (i = 1; i < N; i++) {
    cin >> p;
    ady[p].push_back(i);
    ady[i].push_back(p);
  }
  for (i = 0; i < N; i++) cin >> C[i];
  DFS(0, 0);
  cout << f[0] << "\n";
}
