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 N = 3e5 + 9;
const int mod = 1e9 + 7;
int col[N];
vector<int> ed[N];
long long dp[N][2];
long long rev(long long n) {
  long long r = 1;
  int b = mod - 2;
  while (b > 0) {
    if (b & 1) {
      r = (r * n) % mod;
    }
    n = (n * n) % mod;
    b >>= 1;
  }
  return r;
}
void dfs(int cur, int p) {
  bool leaf = true;
  long long zero = 1;
  for (vector<int>::iterator vit = ed[cur].begin(); vit != ed[cur].end();
       ++vit) {
    if ((*vit) != p) {
      leaf = false;
      dfs(*vit, cur);
      zero = (zero * (dp[*vit][0] + dp[*vit][1])) % mod;
    }
  }
  if (leaf) {
    dp[cur][col[cur]] = 1;
    return;
  }
  if (!col[cur]) {
    dp[cur][0] = zero;
    for (vector<int>::iterator vit = ed[cur].begin(); vit != ed[cur].end();
         ++vit) {
      if ((*vit) != p) {
        dp[cur][1] =
            (dp[cur][1] +
             ((zero * rev(dp[*vit][0] + dp[*vit][1])) % mod) * dp[*vit][1]) %
            mod;
        dp[cur][1] %= mod;
      }
    }
  } else {
    dp[cur][1] = zero;
  }
}
int main() {
  int i, j, n, k;
  while (cin >> n) {
    for (i = 0; i < n; ++i) {
      ed[i].clear();
    }
    for (i = 1; i < n; ++i) {
      cin >> k;
      ed[i].push_back(k);
      ed[k].push_back(i);
    }
    for (i = 0; i < n; ++i) {
      cin >> col[i];
    }
    memset(dp, 0, sizeof(dp));
    dfs(0, -1);
    cout << dp[0][1] << endl;
  }
  return 0;
}
