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.math.BigInteger;
import java.util.ArrayList;
import java.util.Scanner;


public class CFD {
	static final long MOD =1000000007;
	static long [] internal;
	static long [] external;
	static boolean [] black;
	static ArrayList<Integer>[] childs;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int [] tree = new int[N];
		black = new boolean[N];
		childs = new ArrayList[N];
		internal = new long[N];
		external = new long[N];
		for(int i=1;i<N;i++) {
			tree[i]=in.nextInt();
			if(childs[tree[i]]==null)
				childs[tree[i]]=new ArrayList<Integer>(4);
			childs[tree[i]].add(i);
		}
		for(int i=0;i<N;i++) {
			black[i]=(in.nextInt()==1);
		}
		solve(0);
		System.out.println(internal[0]);
	}
	private static void solve(int x) {
		if(childs[x]==null) {
			external[x]=1;
			if(black[x])
				internal[x]=1;
			else
				internal[x]=0;
			return;
		}
		for(int i=0;i<childs[x].size();i++) {
			solve(childs[x].get(i));
		}
		long prodall = 1;
		
		for(int i=0;i<childs[x].size();i++) {
			prodall=(prodall*external[childs[x].get(i)])%MOD;
		}
		if(black[x]) {
			internal[x]=prodall;
			external[x]=prodall;
			return;
		}
		for(int i=0;i<childs[x].size();i++) {
			int t = childs[x].get(i);
			internal[x] = (internal[x]+(prodall*inv(external[t])%MOD*internal[t])%MOD)%MOD;
		}
		external[x]=prodall;
		external[x]=(external[x]+internal[x])%MOD;
	}
	private static long inv(long x) {
		BigInteger t = BigInteger.valueOf(x);
		return t.modInverse(BigInteger.valueOf(MOD)).longValue();
	}

}
