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.io.*;
import java.util.*;

public class F {

	static ArrayList<Integer>[] adj;
	static int a[], dp[][], MOD = (int) 1e9 + 7;

	static void dfs(int u) {
		dp[u][a[u]] = 1;
		for (int v : adj[u]) {
			dfs(v);

			// calc for num=1
			long ans = 0;
			// take edge
			ans = dp[v][1] * 1L * dp[u][0] + dp[v][0] * 1L * dp[u][1];
			ans %= MOD;
			// leave edge
			ans += dp[u][1] * 1L * dp[v][1] % MOD;
			ans %= MOD;
			dp[u][1] = (int) ans;

			// calc for num = 0;

			// take this edge
			ans = dp[u][0] * 1L * dp[v][0] % MOD;
			// leave this edge
			ans += dp[u][0] * 1L * dp[v][1] % MOD;
			dp[u][0] = (int) (ans % MOD);

		}

	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		adj = new ArrayList[n];
		for (int i = 0; i < n; i++)
			adj[i] = new ArrayList();
		for (int i = 1; i < n; i++)
			adj[sc.nextInt()].add(i);
		a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();
		dp = new int[n][2];
		dfs(0);
		out.println(dp[0][1]);
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}