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 = 1000 * 1000 * 1000 + 7;
int power(int b, int n) {
  if (n == 1) {
    return b % MOD;
  } else if (n % 2 == 0) {
    int t = power(b, n / 2);
    return (1LL * t * t) % MOD;
  } else {
    return (1LL * b * power(b, n - 1)) % MOD;
  }
}
int MODdivide(int a, int b) { return (1LL * a * power(b, MOD - 2)) % MOD; }
void dfs(int curr, const vector<vector<int>>& gr, vector<pair<int, int>>& dp,
         vector<int>& color, int parent = -1) {
  int sum = 1;
  for (int next : gr[curr]) {
    if (next == parent) {
      continue;
    }
    dfs(next, gr, dp, color, curr);
    sum = (1LL * sum * (dp[next].first + dp[next].second)) % MOD;
  }
  if (gr[curr].size() == 1 && curr > 0) {
    dp[curr].first = (color[curr] == 0 ? 1 : 0);
    dp[curr].second = (color[curr] == 1 ? 1 : 0);
    return;
  }
  if (color[curr] == 1) {
    dp[curr].first = 0;
    dp[curr].second = sum;
  } else {
    dp[curr].first = sum;
    for (int next : gr[curr]) {
      if (next == parent) {
        continue;
      }
      dp[curr].second = (1LL * dp[curr].second +
                         1LL * dp[next].second *
                             MODdivide(sum, dp[next].first + dp[next].second)) %
                        MOD;
    }
  }
  return;
}
int main() {
  int n;
  cin >> n;
  vector<vector<int>> gr(n);
  for (int i = 0; i < n - 1; ++i) {
    int p;
    cin >> p;
    gr[p].push_back(i + 1);
    gr[i + 1].push_back(p);
  }
  vector<int> color(n);
  for (int i = 0; i < n; ++i) {
    cin >> color[i];
  }
  vector<pair<int, int>> dp(n, {0, 0});
  dfs(0, gr, dp, color);
  cout << dp[0].second << endl;
  return 0;
}
