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;

class Program{
    static void Main(string[] args){
        long M = 1000000007;
        
        string[] str = Console.ReadLine().Split(' ');
        int N = int.Parse(str[0]);
        int C = int.Parse(str[1]);
        str  = Console.ReadLine().Split(' ');
        long[] A = new long[N+1];
        long[] B = new long[N+1];
        for(int i=0;i<N;i++){
            A[i] = long.Parse(str[i]);
        }
        str  = Console.ReadLine().Split(' ');
        for(int i=0;i<N;i++){
            B[i] = long.Parse(str[i]);
        }
        
        int tableSize = 401;
        long[,] table = new long[tableSize,tableSize];
        for(int j=0;j<tableSize;j++){
            table[0,j] = 1;
            for(int i=1;i<tableSize;i++){
                table[i,j] = table[i-1,j] * ((long) j) % M;
            }
        }
        for(int i=0;i<tableSize;i++){
            for(int j=1;j<tableSize;j++){
                table[i,j] += table[i,j-1];
                table[i,j] %= M;
            }
        }
        
        
        
        long[,] dp = new long[N, C+1];
        for(int i=0;i<N;i++){
            for(int j=0;j<C+1;j++){
                dp[i,j] = 0;
            }
        }
        dp[N-1,0]=(table[0,B[N-1]]-table[0,A[N-1]-1])%M;
        for(int j=1;j<C+1;j++){
            dp[N-1,j] = (M+table[j,B[N-1]]-table[j,A[N-1]-1])%M;
        }
        
        for(int i=N-2;i>-1;i--){
            for(int j=0;j<C+1;j++){
                for(int k=0;k<=j;k++){
                    dp[i,j] += ((table[k,B[i]]-table[k,A[i]-1]) * dp[i+1,j-k] % M)% M;
                    dp[i,j] %= M;
                }
            }
        }
        Console.WriteLine(dp[0,C]%M);
    }
}