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 Main {

	static int n;
	static List<List<Integer>> v;
	static int[] color;
	static long[][] dp;
	static final int MOD = 1000000007;

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);

		while (cin.hasNext()) {
			n = cin.nextInt();
			color = new int[n];
			dp = new long[n][2];
			v = new ArrayList<List<Integer>>();
			for (int i = 0; i < n; i++)
				v.add(new ArrayList<Integer>());

			for (int i = 1; i < n; i++) {
				int p = cin.nextInt();
				v.get(i).add(p);
				v.get(p).add(i);
			}
			for (int i = 0; i < n; i++)
				color[i] = cin.nextInt();

			dfs(0, -1);

			System.out.println(dp[0][1]);
		}
		cin.close();
	}

	private static void dfs(int i, int fa) {
		dp[i][color[i]] = 1;
		for (int j : v.get(i)) {
			if (j != fa) {
				dfs(j, i);
				dp[i][1] = (dp[i][1] * (dp[j][0] + dp[j][1]) % MOD + dp[i][0]
						* dp[j][1] % MOD)
						% MOD;
				dp[i][0] = dp[i][0] * (dp[j][0] + dp[j][1]) % MOD;
			}
		}
	}
}
