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
//package codeforces;

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

/**
 * Created by nitin.s on 06/03/16.
 */
public class ApplemanAndTree {
    static List<Integer>[] g;
    static final int P = 1000000007;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        g = new List[n];
        for(int i = 0; i < n; ++i) {
            g[i] = new ArrayList<Integer>(0);
        }
        for(int i = 1; i < n; ++i) {
            g[in.nextInt()].add(i);
        }
        int[] col = new int[n];
        for(int i = 0; i < n; ++i) {
            col[i] = in.nextInt();
        }
        int[][] dp = new int[n][2];
        for(int i = n - 1; i >= 0; --i) {
            dp[i][col[i]] = 1;
            for(int to: g[i]) {
                int new1 = 0;
                int new0 = 0;
                // take
                new1 = (int)((long)dp[i][1] * dp[to][0] % P + (long)dp[i][0] * dp[to][1] % P) % P;
                new0 = (int)((long) dp[i][0] * dp[to][0] % P);
                // skip
                new1 += (int) ((long) dp[i][1] * dp[to][1] % P);
                new1 %= P;
                new0 += (int) ((long) dp[i][0] * dp[to][1] % P);
                new0 %= P;
                dp[i][0] = new0;
                dp[i][1] = new1;
            }
        }
        System.out.println(dp[0][1]);
    }
}
