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 = 1e5 + 55;
vector<int> vec[N];
int m = 1e9 + 7;
long long dp1[N];
long long dp2[N];
int a[N];
void dfs(int u, int parent) {
  dp1[u] = 1;
  dp2[u] = 0;
  for (auto v : vec[u]) {
    if (v == parent) continue;
    dfs(v, u);
    dp2[u] *= dp1[v];
    dp2[u] += dp1[u] * dp2[v];
    dp1[u] *= dp1[v];
    dp1[u] %= m;
    dp2[u] %= m;
  }
  if (a[u] == 1)
    dp2[u] = dp1[u];
  else
    dp1[u] += dp2[u];
  dp1[u] %= m;
  dp2[u] %= m;
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  int x;
  for (int i = 1; i < n; i++) {
    cin >> x;
    vec[i].push_back(x);
    vec[x].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> a[i];
  dfs(0, 0);
  cout << dp2[0] % m << endl;
  return 0;
}
