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 = 3e5;
const long long P = 1e9 + 7;
long long n;
long long col[N], f[N], g[N];
vector<long long> a[N];
long long read(void) {
  long long s = 0, w = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
  return w ? -s : s;
}
void Dfs(long long x, long long fa) {
  col[x] ? f[x] = 1 : g[x] = 1;
  for (long long i = 0; i < a[x].size(); ++i) {
    long long y = a[x][i];
    if (y == fa) continue;
    Dfs(y, x);
    f[x] = (f[x] * (f[y] + g[y]) + g[x] * f[y]) % P;
    g[x] = (g[x] * (f[y] + g[y])) % P;
  }
  return;
}
signed main(void) {
  n = read();
  for (long long i = 1; i < n; ++i) a[read()].push_back(i);
  for (long long i = 0; i < n; ++i) col[i] = read();
  Dfs(0, 0);
  cout << f[0] << endl;
  return 0;
}
