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;
int n, p[100010], x[100010];
vector<int> ch[100010];
long long a[2][100010];
long long modpow(long long q, long long w) {
  long long res = 1;
  q %= 1000000007;
  while (w) {
    if (w & 1) res = res * q % 1000000007;
    q = q * q % 1000000007;
    w >>= 1;
  }
  return res;
}
void dfs(int u) {
  a[0][u] = 1;
  for (int i = 0; i < ch[u].size(); i++) {
    int v = ch[u][i];
    dfs(v);
    a[0][u] *= a[0][v] + a[1][v];
    a[0][u] %= 1000000007;
  }
  if (x[u]) {
    a[1][u] = a[0][u];
    a[0][u] = 0;
    return;
  }
  for (int i = 0; i < ch[u].size(); i++) {
    int v = ch[u][i];
    long long b =
        a[0][u] * modpow(a[0][v] + a[1][v], 1000000007 - 2) % 1000000007;
    b = b * a[1][v] % 1000000007;
    a[1][u] += b;
    a[1][u] %= 1000000007;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> p[i];
    ch[p[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> x[i];
  dfs(0);
  cout << a[1][0] << '\n';
}
