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, lim = 1e6 + 10;
inline long long pwr(long long a, long long n, long long m = mod) {
  long long ans(1);
  while (n) {
    if (n & 1) ans = ans * a % m;
    n >>= 1;
    a = a * a % m;
  }
  return ans;
}
vector<int> t[lim];
int a[lim], x[lim], y[lim], temp;
inline void bfs(int v, int par) {
  bool is_leaf = true;
  for (int ch : t[v])
    if (ch != par) bfs(ch, v), is_leaf = false;
  if (is_leaf) {
    x[v] = 1;
    y[v] = a[v];
    return;
  }
  temp = 1;
  for (int ch : t[v])
    if (ch != par) temp = (long long)temp * x[ch] % mod;
  x[v] = y[v] = temp;
  if (!a[v]) {
    temp = 0;
    for (int ch : t[v])
      if (ch != par)
        temp = (temp + (long long)pwr(x[ch], mod - 2) * y[ch]) % mod;
    y[v] = (long long)y[v] * temp % mod;
    x[v] = (x[v] + y[v]) % mod;
  }
}
int main() {
  int n;
  cin >> n;
  for (long long i = 0, _n = (n - 1); i < _n; i++)
    cin >> temp, t[temp].emplace_back(i + 1), t[i + 1].emplace_back(temp);
  for (long long i = 0, _n = (n); i < _n; i++) cin >> a[i];
  bfs(0, 0);
  cout << y[0];
  return 0;
}
