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>
const int mod = (int)1e9 + 7;
int add(int a, int b) { return (a += b) >= mod ? a -= mod : a; }
int sub(int a, int b) { return add(a, mod - b); }
int mul(int a, int b) { return int(1LL * a * b % mod); }
int pow(int a, int64_t n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
int inv(int a) { return pow(a, mod - 2); }
int divmod(int a, int b) { return mul(a, inv(b)); }
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  for (int n; std::cin >> n;) {
    std::vector<std::vector<int> > next(n);
    for (int u = 1, p; u < n; ++u) {
      std::cin >> p;
      next[p].push_back(u);
      next[u].push_back(p);
    }
    std::vector<int> color(n);
    for (int i = 0; i < n; ++i) {
      std::cin >> color[i];
    }
    std::vector<int> cnt0(n, -1), cnt1(n);
    std::function<void(int, int)> dfs = [&](int u, int p) {
      if (color[u] == 1) {
        cnt0[u] = 0;
        cnt1[u] = 1;
        return;
      }
      cnt0[u] = 1;
      cnt1[u] = 0;
      for (int v : next[u]) {
        if (v == p) {
          continue;
        }
        dfs(v, u);
        cnt0[u] = mul(cnt0[u], add(mul(cnt0[v], 1), cnt1[v]));
        cnt1[u] = add(cnt1[u], divmod(cnt1[v], add(mul(cnt0[v], 1), cnt1[v])));
      }
      cnt1[u] = mul(cnt1[u], cnt0[u]);
    };
    int answ = 1;
    for (int u = 0; u < n; ++u) {
      if (cnt0[u] == -1 && color[u] == 0) {
        dfs(u, -1);
        answ = mul(answ, cnt1[u]);
      }
    }
    std::cerr && false && std::cerr << std::string(40, '-') << std::endl;
    for (int u = 0; u < n; ++u) {
      std::cerr && false &&
          std::cerr << "u = " << u << ", cnt0 = " << cnt0[u]
                    << ", cnt1 = " << cnt1[u] << std::endl;
    }
    std::cerr && false && std::cerr << std::string(40, '-') << std::endl;
    std::cout << answ << std::endl;
  }
  return 0;
}
