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 long long N = 1e5 + 1;
long long mod = 1e9 + 7;
vector<long long> adj[N];
long long col[N];
long long dp[N][2];
void dfs(long long src, long long par) {
  dp[src][0] = !col[src];
  dp[src][1] = col[src];
  for (auto &x : adj[src]) {
    if (x == par) continue;
    dfs(x, src);
    dp[src][1] =
        dp[src][1] * dp[x][1] + dp[src][1] * dp[x][0] + dp[src][0] * dp[x][1];
    dp[src][1] %= mod;
    dp[src][0] = dp[src][0] * dp[x][0] + dp[src][0] * dp[x][1];
    dp[src][0] %= mod;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    p++;
    adj[i].push_back(p);
    adj[p].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> col[i];
  }
  dfs(1, -1);
  cout << dp[1][1];
}
