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;
constexpr long long mod = 1000000007;
pair<long long, long long> dfs(int v, const vector<vector<int>> &children,
                               const vector<int> &black) {
  long long zero = 0;
  long long one = 0;
  (black[v] ? one : zero) = 1;
  for (const auto &child : children[v]) {
    const auto result = dfs(child, children, black);
    long long t_zero = (zero * result.second + zero * result.first) % mod;
    one =
        (one * result.second + zero * result.second + one * result.first) % mod;
    zero = t_zero;
  }
  return {zero, one};
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<vector<int>> children(n);
  for (int i = 1; i < n; ++i) {
    int p;
    cin >> p;
    children[p].emplace_back(i);
  }
  vector<int> black(n);
  for (auto &e : black) cin >> e;
  const auto ans = dfs(0, children, black);
  cout << ans.second << endl;
  return EXIT_SUCCESS;
}
