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;
int n;
vector<int> g[100000];
int b[100000];
vector<pair<int, int> > t[100000];
long long P(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = (ret * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return ret;
}
long long D(long long a, long long b) { return (a * P(b, MOD - 2)) % MOD; }
pair<int, int> dfs(int v) {
  if (b[v]) {
    long long ret = 1;
    for (int to : g[v]) {
      pair<int, int> tmp = dfs(to);
      ret = (ret * (tmp.first + tmp.second)) % MOD;
    }
    return make_pair(0, (int)ret);
  } else {
    long long ret = 1, ret2 = 0;
    for (int to : g[v]) {
      pair<int, int> tmp = dfs(to);
      t[v].push_back(tmp);
      ret = (ret * (tmp.first + tmp.second)) % MOD;
    }
    for (pair<int, int> z : t[v])
      ret2 = (ret2 + D(ret, z.first + z.second) * z.second) % MOD;
    return make_pair((int)ret, (int)ret2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x;
    cin >> x;
    g[x].push_back(i);
  }
  for (int i = 0; i < (int)(n); i++) cin >> b[i];
  cout << dfs(0).second << '\n';
  return 0;
}
