462_D. Appleman and Tree_17196
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 = 1000000007;
const int MAXN = 100010;
int N, c[MAXN], f[MAXN][2];
vector<int> graph[MAXN];
void dfs(int u, int p) {
  f[u][0] = 1;
  f[u][1] = 0;
  for (int i = 0; i < graph[u].size(); ++i) {
    int v = graph[u][i];
    if (v != p) {
      dfs(v, u);
      f[u][1] = (long long)f[u][1] * f[v][0] % MOD;
      f[u][1] = (f[u][1] + (long long)f[u][0] * f[v][1]) % MOD;
      f[u][0] = (long long)f[u][0] * f[v][0] % MOD;
    }
  }
  if (c[u] == 1)
    f[u][1] = f[u][0];
  else
    f[u][0] = (f[u][0] + f[u][1]) % MOD;
}
int main() {
  int v;
  cin >> N;
  for (int u = 0; u < N - 1; ++u) {
    cin >> v;
    graph[u + 1].push_back(v);
    graph[v].push_back(u + 1);
  }
  for (int i = 0; i < N; ++i) cin >> c[i];
  dfs(0, -1);
  cout << f[0][1] << endl;
  return 0;
}
