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
import java.util.Scanner;

public class TestD {
    final static long MOD = (int)1e9 + 7;

    public static void main (String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] parent = new int[n];
        for (int i = 1; i < n; i++) {
                parent[i] = sc.nextInt();
        }
        long[][] colors = new long[n][2];
        for (int i = 0; i < n; ++ i) {
           colors[i][sc.nextInt()] = 1;
        }
            for (int v = n-1; v >=1; v--) {
                int p = parent[v];
                colors[p] = new long[]{(colors[p][0] * colors[v][0] % MOD + colors[p][0] * colors[v][1] % MOD) % MOD,
                                     (colors[p][0] * colors[v][1] % MOD + colors[p][1] * colors[v][0] % MOD + colors[p][1] * colors[v][1] % MOD) % MOD};
            }
            System.out.println(colors[0][1]);
    }

}