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 + 10;
const int M = 1e9 + 7;
enum Color { W, B };
int n, x[N], p[N], f[N][2];
vector<int> g[N];
void Dfs(int u) {
  f[u][W] = (x[u] == W);
  f[u][B] = (x[u] == B);
  for (auto v : g[u]) {
    int t[2];
    memcpy(t, f[u], sizeof(t));
    Dfs(v);
    t[W] = (x[u] == W ? ((((1ll * (f[u][W]) * (f[v][W]) % M)) +
                          ((1ll * (f[u][W]) * (f[v][B]) % M))) %
                         M)
                      : 0);
    t[B] = ((((1ll * (f[u][B]) * (f[v][W]) % M)) +
             ((1ll * (f[u][B]) * (f[v][B]) % M))) %
            M);
    if (x[u] == W) t[B] = (((t[B]) + ((1ll * (f[u][W]) * (f[v][B]) % M))) % M);
    memcpy(f[u], t, sizeof(t));
  }
}
int main() {
  cin.sync_with_stdio(0);
  while (cin >> n) {
    for (int i = 0; i < n; i++) g[i].clear();
    for (int i = 0; i < n - 1; i++) {
      int u;
      cin >> u;
      int v = i + 1;
      g[u].push_back(v);
    }
    for (int i = 0; i < n; i++) cin >> x[i];
    Dfs(0);
    cout << f[0][B] << endl;
  }
  return 0;
}
