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 = 1e9 + 7;
vector<long long> v[112345];
long long n, p[112345], f[112345][2];
bool vis[112345];
void dfs(long long x) {
  vis[x] = 1;
  for (long long i = 0; i < v[x].size(); i++) {
    long long dir = v[x][i];
    if (vis[dir]) continue;
    dfs(dir);
    f[x][1] =
        (f[x][1] * (f[dir][1] + f[dir][0]) % N + f[x][0] * f[dir][1] % N) % N;
    f[x][0] = (f[x][0] * (f[dir][1] + f[dir][0])) % N;
  }
  vis[x] = 0;
}
signed main() {
  scanf("%lld", &n);
  long long u;
  for (long long i = 1; i < n; i++) {
    scanf("%lld", &u);
    v[i].push_back(u);
    v[u].push_back(i);
  }
  for (long long i = 0; i < n; i++) {
    scanf("%lld", &p[i]);
    f[i][p[i]] = 1;
  }
  dfs(0);
  printf("%lld", f[0][1] % N);
  return 0;
}
