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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B {
    static ArrayList<Integer>[] tree;
    static int[][][] dp;
    static int[] color;

    static long MOD = 1000000007;

    public static int f(int i, int j, int need, int p) {
        if ((need == 0 && color[i] == 1 && j == 0) || need < 0)
            return 0;
        if (need == 1 && color[i] == 1)
            need = 0;
        if (j == tree[i].size())
            return need == 0 ? 1 : 0;
        if (dp[i][j][need] != -1)
            return dp[i][j][need];
        int node = tree[i].get(j);
        if (node == p)
            return dp[i][j][need] = f(i, j + 1, need, p);
        // connect or not connect
        long x = (f(node, 0, 1, i) * 1L * f(i, j + 1, need - 1, p) + f(node, 0,
                0, i) * 1L * f(i, j + 1, need, p))
                % MOD;
        long y = (f(node, 0, 1, i) * 1L * f(i, j + 1, need, p)) % MOD;
        return dp[i][j][need] = (int) ((x + y) % MOD);
    }

    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer strtok;
        strtok = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(strtok.nextToken());
        tree = new ArrayList[n];
        for (int i = 0; i < n; i++)
            tree[i] = new ArrayList<>();
        strtok = new StringTokenizer(in.readLine());
        for (int i = 0; i < n - 1; i++) {
            int pi = Integer.parseInt(strtok.nextToken());
            tree[(i + 1)].add(pi);
            tree[pi].add(i + 1);
        }
        dp = new int[n][][];
        for (int i = 0; i < n; i++) {
            dp[i] = new int[tree[i].size()][2];
            for (int[] j : dp[i])
                Arrays.fill(j, -1);
        }
        color = new int[n];
        strtok = new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++)
            color[i] = Integer.parseInt(strtok.nextToken());
        System.out.println(f(0, 0, 1, -1));

    }
}
