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 V = 100100;
const int P = 1000000007;
struct Edge {
  int v, ne;
} e[V * 2];
int p[V], K;
void add(int u, int v) {
  e[K].v = v;
  e[K].ne = p[u];
  p[u] = K++;
}
int col[V], dp[V][2];
void dfs(int u) {
  dp[u][0] = dp[u][1] = 0;
  if (col[u] == 1)
    dp[u][1] = 1;
  else
    dp[u][0] = 1;
  for (int i = p[u]; i != -1; i = e[i].ne) {
    int v = e[i].v;
    dfs(v);
    int tmp0 = (long long)dp[u][0] * (dp[v][0] + dp[v][1]) % P;
    int tmp1 = ((long long)(dp[u][0] + dp[u][1]) * dp[v][1] % P +
                (long long)dp[u][1] * dp[v][0] % P) %
               P;
    dp[u][0] = tmp0;
    dp[u][1] = tmp1;
  }
}
int n, x;
int main() {
  while (~scanf("%d", &n)) {
    memset(p, -1, sizeof(p));
    K = 0;
    for (int i = 1; i < n; i++) {
      scanf("%d", &x);
      add(x, i);
    }
    for (int i = 0; i < n; i++) scanf("%d", &col[i]);
    dfs(0);
    printf("%d\n", dp[0][1]);
  }
  return 0;
}
