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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const long long P = 1000000007LL;
const int maxN = 110000;
vector<int> g[maxN];
int n;
int c[maxN];
long long d[2][maxN];
long long dp[maxN][2];
void dfs(int v) {
  for (int i = 0; i < g[v].size(); ++i) {
    dfs(g[v][i]);
  }
  if (g[v].size() == 0) {
    d[c[v]][v] = 1;
    return;
  }
  for (int i = 0; i <= g[v].size(); ++i) {
    dp[i][0] = dp[i][1] = 0;
  }
  dp[0][c[v]] = 1;
  for (int i = 0; i < g[v].size(); ++i) {
    for (int j = 0; j < 2; ++j) {
      if (dp[i][j] == 0) {
        continue;
      }
      {
        for (int k = 0; k < 2; ++k) {
          if (j + k < 2) {
            dp[i + 1][j + k] += dp[i][j] * d[k][g[v][i]];
            dp[i + 1][j + k] %= P;
          }
        }
      }
      {
        dp[i + 1][j] += dp[i][j] * d[1][g[v][i]];
        dp[i + 1][j] %= P;
      }
    }
  }
  d[0][v] = dp[g[v].size()][0];
  d[1][v] = dp[g[v].size()][1];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int v;
    scanf("%d", &v);
    g[v].push_back(i + 1);
  }
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
    cnt += c[i];
  }
  if (cnt == 0) {
    printf("0\n");
    return 0;
  }
  dfs(0);
  cout << d[1][0] << endl;
  return 0;
}
