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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.TreeMap;




public class Main {
	  static int MOD=1000000007;
	  static long p[][]=new long[405][405];
	  static long sum[][]=new long[405][405];
	  static long dp[][]=new long[405][405];
	  static long a[]=new long[405];
	  static long b[]=new long[405];
      public static void main(String[] args) {
    	  Scanner in=new Scanner( new BufferedReader(new InputStreamReader(System.in)))	;
    
    	  
    	  for(int i=1;i<=404;i++)//p[i][j]=i^j
    	  {
    		  p[i]= new long[405];
    		  p[i][0]=1;
    		  for(int j=1;j<=404;j++)
    		  {
    			  p[i][j]=(p[i][j-1]*i)%MOD;
    		  }
    	  }
    	 
    	  for(int i=1;i<=404;i++)//sum[i][j] 1^j+...+i^j
    	  {
    		  sum[i]=new long[405];
    		 
    		  for(int j=0;j<=404;j++)
    		  {
    			  sum[0][j]=0;
    			  sum[i][j]=(sum[i-1][j]%MOD+p[i][j]%MOD)%MOD;
    		  }
    	  }
    	  int n=in.nextInt();
    	  int m=in.nextInt();
    	  
    	  for(int i=1;i<=n;i++)
    	  {
    		  a[i]=in.nextLong();
    	  }
    	  for(int i=1;i<=n;i++)
    	  {
    		  b[i]=in.nextLong();
    	  }
    	  
    	  dp[0][0]=1;
    	  for(int i=1;i<=n;i++)
    	  {
    		  for(int j=0;j<=m;j++)
    		  {
    			  for(int k=0;k<=j;k++)
    			  {
    				  long temp=(sum[(int) b[i]][k]%MOD-sum[(int) (a[i]-1)][k]%MOD+MOD)%MOD;
    				  dp[i][j]=(dp[i][j]%MOD+dp[i-1][j-k]*temp%MOD)%MOD;
    			  }
    		  }
    	  }
    	  System.out.println(dp[n][m]);
    	  return;
      }
}