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.Arrays;
import java.util.Scanner;

public class CF461B {
	static final int MOD = 1000000007;
	static long[][] f;
	static boolean[] visited;
	static int[] color;
	static ArrayList<ArrayList<Integer>> adjList;
	
	static void dfs(int u) {
		int numAdj = adjList.get(u).size();
		visited[u] = true;
		f[u][0] = 1; f[u][1] = 0;
		for(int i=0; i<numAdj; i++) {
			int v = adjList.get(u).get(i);
			if (!visited[v]) {
				dfs(v);
				f[u][1] = (((f[u][1]*f[v][0])%MOD) + (f[u][0]*f[v][1])%MOD)%MOD;
				f[u][0] = (f[u][0]*f[v][0])%MOD;
			}	
		}
		if (color[u] == 1) f[u][1] = f[u][0];
		else f[u][0] = (f[u][0]+f[u][1])%MOD;
	}
	
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		f = new long[N+1][2];
		visited = new boolean[N+1];
		Arrays.fill(visited, false);
		adjList = new ArrayList<ArrayList<Integer>>();
		for(int i=0; i<N; i++) adjList.add(new ArrayList<Integer>());
		for(int i=0; i<N-1; i++) {
			int a = in.nextInt();
			adjList.get(i+1).add(a);
			adjList.get(a).add(i+1);
		}
		color = new int[N+1];
		for(int i=0; i<N; i++) color[i] = in.nextInt();
		dfs(0);
		System.out.println(f[0][1]);
		in.close();
	}

}
