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 pow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = pow(a, b / 2);
  res *= res;
  res %= 1000000007;
  if (b % 2 == 1) res *= a;
  return res % 1000000007;
}
long long rev(long long x) { return pow(x, 1000000007 - 2); }
long long color[108000], p[108000];
long long dp[108000][2];
vector<int> edge[108000];
long long dfs(int x) {
  for (int i = 0; i < edge[x].size(); i++) {
    int to = edge[x][i];
    dfs(to);
  }
  if (color[x] == 1) {
    dp[x][0] = 0;
    dp[x][1] = 1;
    for (int i = 0; i < edge[x].size(); i++) {
      int to = edge[x][i];
      dp[x][1] *= dp[to][0] + dp[to][1];
      dp[x][1] %= 1000000007;
    }
  } else {
    dp[x][0] = 1;
    dp[x][1] = 0;
    for (int i = 0; i < edge[x].size(); i++) {
      int to = edge[x][i];
      dp[x][0] *= dp[to][0] + dp[to][1];
      dp[x][0] %= 1000000007;
    }
    for (int i = 0; i < edge[x].size(); i++) {
      int to = edge[x][i];
      dp[x][1] +=
          (dp[x][0] * dp[to][1]) % 1000000007 * rev(dp[to][0] + dp[to][1]);
      dp[x][1] %= 1000000007;
    }
  }
  return dp[x][1];
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    cin >> p[i];
    edge[p[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> color[i];
  cout << dfs(0) << endl;
  return 0;
}
