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: Accepted)

import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;

public class Main implements Runnable {
	public static void main(String[] args) {
		new Thread(null, new Main(), "", Runtime.getRuntime().maxMemory()).start();
		// new Main().run();
	}

	final long MODULO = (long) (1e9 + 7);
	int n, c;
	int[] a, b;

	public void run() {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		c = sc.nextInt();
		a = new int[n];
		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();
		}

		long[][] pow = new long[c + 1][401];
		Arrays.fill(pow[0], 1);
		for (int i = 1; i <= c; ++i) {
			for (int j = 0; j < pow[i].length; ++j) {
				pow[i][j] = pow[i - 1][j] * j % MODULO;
			}
		}
		long[][] sum = new long[c + 1][401];
		for (int i = 0; i <= c; ++i) {
			for (int j = 0; j < pow[i].length; ++j) {
				sum[i][j] += (j > 0 ? sum[i][j - 1] : 0) + pow[i][j];
				if (sum[i][j] >= MODULO)
					sum[i][j] -= MODULO;
			}
		}

		long[][] g = new long[c + 1][n];
		for (int i = 0; i <= c; ++i) {
			for (int j = 0; j < n; ++j) {
				g[i][j] = sum[i][b[j]] - (a[j] > 0 ? sum[i][a[j] - 1] : 0);
				g[i][j] = (g[i][j] + MODULO) % MODULO;
			}
		}

		long[][] f = new long[n][c + 1];
		for (int i = 0; i <= c; ++i) {
			f[0][i] = g[i][0];
		}

		for (int i = 1; i < n; ++i) {
			for (int preC = 0; preC <= c; ++preC) {
				for (int curC = 0; preC + curC <= c; ++curC) {
					f[i][preC + curC] += f[i - 1][preC] * g[curC][i] % MODULO;
					f[i][preC + curC] %= MODULO;
				}
			}
		}

		System.out.println(f[n - 1][c] % MODULO);
	}

	static void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}