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;
int n, x, kk, dp[2000005][2], col[2000005], head[2000005];
struct Tree {
  int nxt, to;
} e[2000005];
inline void link(int x, int y) {
  e[++kk].nxt = head[x];
  e[kk].to = y;
  head[x] = kk;
}
void dfs1(int u, int fa) {
  dp[u][col[u]] = 1;
  dp[u][col[u] ^ 1] = 0;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    dp[u][1] = 1ll *
               (1ll * dp[u][1] * (dp[v][0] + dp[v][1]) % 1000000007 +
                1ll * dp[u][0] * dp[v][1] % 1000000007) %
               1000000007;
    dp[u][0] = 1ll *
               (1ll * dp[u][0] * dp[v][0] % 1000000007 +
                1ll * dp[u][0] * dp[v][1] % 1000000007) %
               1000000007;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &x);
    x++;
    link(x, i);
  }
  for (int i = 1; i <= n; i++) scanf("%d", &col[i]);
  dfs1(1, -1);
  printf("%d\n", (dp[1][1]) % 1000000007);
  return 0;
}
