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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
class Solve{
    int N;
    int C;
    int[] a;
    int[] b;
    public Solve(){}
    public static int Main(){
        new Solve().Calc();
        return 0;
    }
    void Calc(){
        getinput();
        long[,] pow = new long[401,C+1];
        long[,] powsum = new long[401,C+1];
        long[,] dp = new long[N+1,C+1];
        for(int i=0;i<401;i++){
            for(int j=0;j<C+1;j++){
                if(j == 0){
                    pow[i,j] = 1;
                }
                else{
                    pow[i,j] = (pow[i,j-1]*i)%Define.mod;
                }
            }
        }
        for(int i=0;i<401;i++){
            for(int j=0;j<C+1;j++){
                if(i == 0){
                    powsum[i,j] = pow[i,j]; 
                }
                else{
                    powsum[i,j] = (pow[i,j]+powsum[i-1,j])%Define.mod;
                }
            }
        }
        for(int i=0;i<N+1;i++){
            for(int j=0;j<C+1;j++){
                if(i == 0 && j == 0){
                    dp[i,j] = 1;
                }
                else if(j == 0){
                    dp[i,j] = (b[i-1]-a[i-1]+1)*dp[i-1,j]%Define.mod;
                }
                else if(i == 0){
                    dp[i,j] = 0;
                }
                else{
                    long count = dp[i-1,j]*(b[i-1]-a[i-1]+1)%Define.mod;
                    for(int k=1;k<j+1;k++){
                        count = (count+((powsum[b[i-1],k]-powsum[a[i-1]-1,k])%Define.mod*dp[i-1,j-k]%Define.mod))%Define.mod; 
                    }
                    dp[i,j] = count;
                }
            }
        }
        Console.WriteLine(dp[N,C]);
        Define.DebugPrint(dp);
    }
    void getinput(){
        string[] str = Console.ReadLine().Split(' ');
        N = int.Parse(str[0]);
        C = int.Parse(str[1]);
        a = new int[N];
        b = new int[N];
        str = Console.ReadLine().Split(' ');
        for(int i=0;i<N;i++){
            a[i] = int.Parse(str[i]);
        }
        str = Console.ReadLine().Split(' ');
        for(int i=0;i<N;i++){
            b[i] = int.Parse(str[i]);
        }
    }    
}
public static class Define{
    public const long mod = 1000000000;
    public static void DebugPrint(long[,] k){
        for(int i=0;i<k.GetLength(0);i++){
            for(int j=0;j<k.GetLength(1);j++){
                Console.Write(k[i,j]+" ");
            }
            Console.WriteLine();
        }
    }
}