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;
vector<int> adj[100005];
int n, num, colors[100005];
long long int dp0[100005], dp1[100005];
void dfs(int v, int pv) {
  dp0[v] = 1 - colors[v];
  dp1[v] = colors[v];
  for (auto u : adj[v]) {
    if (u == pv) continue;
    dfs(u, v);
    long long int temp = dp1[v];
    dp1[v] =
        (temp * dp0[u]) % MOD + (dp1[u] * temp) % MOD + (dp0[v] * dp1[u]) % MOD;
    if (dp1[v] >= MOD) dp1[v] %= MOD;
    temp = dp0[v];
    dp0[v] = (temp * dp0[u]) % MOD + (dp1[u] * temp) % MOD;
    if (dp0[v] >= MOD) dp0[v] %= MOD;
  }
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> num;
    adj[i].push_back(num + 1);
    adj[num + 1].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> colors[i];
  dfs(1, 0);
  cout << dp1[1] << endl;
  return 0;
}
