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 = 100001, MOD = 1e9 + 7;
int n, clr[N];
long long dp[N][2];
vector<vector<int> > g;
void dfs(int u, int p) {
  if (clr[u] == 0) {
    dp[u][0] = 1;
    dp[u][1] = 0;
    for (int i = 0; i < g[u].size(); i++)
      if (g[u][i] != p) {
        dfs(g[u][i], u);
        dp[u][1] = (dp[u][1] * dp[g[u][i]][0]) % MOD;
        dp[u][1] = (dp[u][1] + (dp[g[u][i]][1] * dp[u][0]) % MOD) % MOD;
        dp[u][0] = (dp[u][0] * dp[g[u][i]][0]) % MOD;
      }
  } else {
    dp[u][1] = 1;
    for (int i = 0; i < g[u].size(); i++)
      if (g[u][i] != p) {
        dfs(g[u][i], u);
        dp[u][1] = (dp[u][1] * dp[g[u][i]][0]) % MOD;
      }
  }
  dp[u][0] = (dp[u][0] + dp[u][1]) % MOD;
}
int main() {
  std::ios::sync_with_stdio(false);
  scanf("%d", &n);
  g.resize(n);
  for (int i = 1; i < n; i++) {
    int p;
    scanf("%d", &p);
    g[i].push_back(p);
    g[p].push_back(i);
  }
  for (int i = 0; i < n; i++) scanf("%d", clr + i);
  dfs(0, -1);
  printf("%d\n", dp[0][1]);
  return 0;
}
