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

public class D {
    static List<Integer> e[];
    static int color[];
    static boolean mark[];
    static int n;
    static long dp[];
    static long bnum[];
    static final long mod = 1000000007;
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        n = cin.nextInt();
        e = new List[n];
        color = new int[n];
        mark = new boolean[n];
        dp = new long[n];
        bnum = new long[n];

        for(int i=0;i<n;i++){
            e[i] = new ArrayList<Integer>();
            mark[i] = false;
        }
        int l = 0;
        for(int i=0;i<n-1;i++){
            l = cin.nextInt();
            add(i+1,l);
        }
        for(int i=0;i<n;i++){
            color[i] = cin.nextInt();
        }
        
        System.out.println(dfs(0));

    }
    public static void add(int a,int b){
        e[a].add(b);
        e[b].add(a);
    }
    public static long dfs(int x){
        if(color[x] == 1){
            dp[x] = 1;
            bnum[x] = 0;
        }
        else{
            bnum[x] = 1;
            dp[x] = 0;
        }
        mark[x] = true;
        for(Integer node : e[x]){
            if(!mark[node]){
                dfs(node);
                dp[x] = (dp[x]*bnum[node] + dp[node]*bnum[x] + dp[x] * dp[node] )%mod;
                bnum[x] = bnum[x] *(bnum[node] + dp[node]) %mod;
            }
        }
        return dp[x];
    }
}
