Problem:
Score : 800 points

Problem Statement

12:17 (UTC): The sample input 1 and 2 were swapped. The error is now fixed. We are very sorry for your inconvenience.

There are N children in AtCoder Kindergarten, conveniently numbered 1 through N. Mr. Evi will distribute C indistinguishable candies to the children.

If child i is given a candies, the child's happiness will become x_i^a, where x_i is the child's excitement level. The activity level of the kindergarten is the product of the happiness of all the N children.

For each possible way to distribute C candies to the children by giving zero or more candies to each child, calculate the activity level of the kindergarten. Then, calculate the sum over all possible way to distribute C candies. This sum can be seen as a function of the children's excitement levels x_1,..,x_N, thus we call it f(x_1,..,x_N).

You are given integers A_i,B_i (1≦i≦N). Find modulo 10^9+7.

Constraints

1≦N≦400

1≦C≦400

1≦A_i≦B_i≦400 (1≦i≦N)

Partial Score

400 points will be awarded for passing the test set satisfying A_i=B_i (1≦i≦N).

Input

The input is given from Standard Input in the following format:

N C
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

Print the value of modulo 10^9+7.

Sample Input 1

2 3
1 1
1 1

Sample Output 1

4

This case is included in the test set for the partial score, since A_i=B_i.
We only have to consider the sum of the activity level of the kindergarten where the excitement level of both child 1 and child 2 are 1 (f(1,1)).

If child 1 is given 0 candy, and child 2 is given 3 candies, the activity level of the kindergarten is 1^0*1^3=1.

If child 1 is given 1 candy, and child 2 is given 2 candies, the activity level of the kindergarten is 1^1*1^2=1.

If child 1 is given 2 candies, and child 2 is given 1 candy, the activity level of the kindergarten is 1^2*1^1=1.

If child 1 is given 3 candies, and child 2 is given 0 candy, the activity level of the kindergarten is 1^3*1^0=1.

Thus, f(1,1)=1+1+1+1=4, and the sum over all f is also 4.

Sample Input 2

1 2
1
3

Sample Output 2

14

Since there is only one child, child 1's happiness itself will be the activity level of the kindergarten. Since the only possible way to distribute 2 candies is to give both candies to child 1, the activity level in this case will become the value of f.

When the excitement level of child 1 is 1, f(1)=1^2=1.

When the excitement level of child 1 is 2, f(2)=2^2=4.

When the excitement level of child 1 is 3, f(3)=3^2=9.

Thus, the answer is 1+4+9=14.

Sample Input 3

2 3
1 1
2 2

Sample Output 3

66

Since it can be seen that f(1,1)=4 , f(1,2)=15 , f(2,1)=15 , f(2,2)=32, the answer is 4+15+15+32=66.

Sample Input 4

4 8
3 1 4 1
3 1 4 1

Sample Output 4

421749

This case is included in the test set for the partial score.

Sample Input 5

3 100
7 6 5
9 9 9

Sample Output 5

139123417

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
	static long pow(int a, int n, int mod) {
		if(n == 0) return 1;
		
		long m = pow(a, n/2, mod);
		if(n%2 == 0) return (m*m)%mod;
		else return (((m*m)%mod)*a)%mod;
	}
	
	static ArrayList<ArrayList<Integer>> comb(int n, int k) {
		ArrayList<ArrayList<Integer>> lst = new ArrayList<ArrayList<Integer>>();
		if(n == 1) {
			ArrayList<Integer> elem = new ArrayList<Integer>();
			elem.add(k);
			lst.add(elem);
			return lst;
		} else {
			for(int i = 0; i <= k; i++) {
				ArrayList<ArrayList<Integer>> suf = comb(n-1, k-i);
				for(int j = 0; j < suf.size(); j++) {
					ArrayList<Integer> elem = new ArrayList<Integer>();
					elem.add(i);
					for(int l = 0; l < n-1; l++) {
						elem.add(suf.get(j).get(l));
					}
					lst.add(elem);
				}
			}
			return lst;
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int c = sc.nextInt();
		int[] a = new int[n];
		int[] b = new int[n];
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		for(int i = 0; i < n; i++) {
			b[i] = sc.nextInt();
			if(a[i] != b[i]) {
				System.out.println(0);
				return;
			}
		}
		
		sc.close();
		
		ArrayList<ArrayList<Integer>> combination = comb(n, c);
		
		int MOD = 1000000007;
		int ans = 0;
		for(int i = 0; i < combination.size(); i++) {
			int subans = 1;
			for(int j = 0; j < n; j++) {
				subans *= pow(a[j], combination.get(i).get(j), MOD);
			}
			ans += subans;
			ans %= MOD;
		}
		
		System.out.println(ans);
	}
	
}