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;
long long int dp[2][200000];
int p[200000];
vector<int> v[200000];
int col[200000];
long long int modexp(long long int a, long long int b) {
  long long int ret = 1;
  while (b) {
    if ((b & 1)) {
      ret *= a;
      ret %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    b >>= 1;
  }
  return ret;
}
long long int getans(int use_black, int cnode, int par = -1) {
  long long int ret = dp[use_black][cnode];
  if (ret == -1) {
    int n = v[cnode].size(), i;
    long long int &ret1 = dp[0][cnode];
    long long int &ret2 = dp[1][cnode];
    ret1 = 1;
    ret2 = 0;
    for (i = 0; i < n; i++) {
      int child = v[cnode][i];
      if (child == par) continue;
      ret1 *= (getans(1, child, cnode) + getans(0, child, cnode)) % 1000000007;
      ret1 %= 1000000007;
    }
    if (col[cnode] == 0) {
      for (i = 0; i < n; i++) {
        int child = v[cnode][i];
        if (child == par) continue;
        long long int mul =
            (getans(1, child, cnode) *
             modexp(getans(0, child, cnode) + getans(1, child, cnode),
                    1000000007 - 2)) %
            1000000007;
        ret2 += (mul * ret1) % 1000000007;
        ret2 %= 1000000007;
      }
    } else {
      ret2 = ret1;
      ret1 = 0;
    }
  }
  return dp[use_black][cnode];
}
int main() {
  memset(dp, -1, sizeof(dp));
  int i, n, x;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> x;
    v[i + 1].push_back(x);
    v[x].push_back(i + 1);
  }
  for (i = 0; i < n; i++) cin >> col[i];
  cout << getans(1, 0) << endl;
  return 0;
}
