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.Scanner;

public class Main {
	public static int N;
	public static ArrayList<ArrayList<Integer>> edges;
	public static long[][] dp;
	public static final long MOD = (long) (1e9 + 7);

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		N = s.nextInt();
		edges = new ArrayList<ArrayList<Integer>>();
		for (int i = 0; i < N; i++) {
			edges.add(new ArrayList<Integer>());
		}
		for (int i = 1; i < N; i++) {
			int a = s.nextInt();
			edges.get(a).add(i);
			edges.get(i).add(a);
		}
		dp = new long[N][2];
		for (int i = 0; i < N; i++) {
			int a = s.nextInt();
			dp[i][a] = 1;
		}
		dfs(0, -1);
		System.out.println(dp[0][1]);
	}

	public static void dfs(int u, int p) {
		for (int v : edges.get(u)) {
			if (v == p)
				continue;
			dfs(v, u);
			dp[u][1] = (dp[u][1] * ((dp[v][0] + dp[v][1]) % MOD) % MOD + dp[u][0] * dp[v][1] % MOD) % MOD;
			dp[u][0] = (dp[u][0] * (dp[v][0] + dp[v][1]) % MOD) % MOD;
		}
	}
}