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: Time Limit Exceeded)

using System;

public class AtCoder
{
    public static void Main()
    {
        string[] s;
        s = NextS();

        int n = int.Parse(s[0]);
        int c = int.Parse(s[1]);
        const int p = 1000000007;

        int[] a = new int[n];
        int[] b = new int[n];
        s = NextS();
        for (int i = 0; i < n; i++) a[i] = int.Parse(s[i]);
        s = NextS();
        for (int i = 0; i < n; i++) b[i] = int.Parse(s[i]);

        Mod mod = new Mod(410);

        long[][] x = new long[n][];
        for (int i = 0; i < n; i++) {
            x[i] = new long[c + 1];
            for (int j = 0; j <= c; j++)
            {
                for(int k=a[i]; k<=b[i]; k++)
                {
                    x[i][j] += mod.Pow(k, j);
                    x[i][j] %= p;
                } 
            }
        }

        long[][] dp = new long[n][];
        dp[0] = new long[c + 1];
        for (int j = 0; j <= c; j++)
        {
            dp[0][j] = x[0][j];
        }
        for (int i = 1; i < n; i++)
        {
            dp[i] = new long[c + 1];
            for(int j=0; j<=c; j++)
            {
                for(int k=0; k<=j; k++)
                {
                    dp[i][j] += dp[i - 1][j - k] * x[i][k];
                    dp[i][j] %= p;
                }               
            }
        }

        Console.WriteLine(dp[n-1][c]);
    }

    public static void DebugWL(object x) { System.Diagnostics.Debug.WriteLine(x); }
    public static string[] NextS() { return Console.ReadLine().Trim().Split(' '); }
}

// Mod(int n)
// int[] fac: factors (1-indexed)
// int n: the length of fac

// int .Ncr(n,r): nCr
// int .Fac(n): n!%p
// int .Pow(n,m): n^m%p

public class Mod
{
    const int p = 1000000007;
    int[] fac;

    public Mod(int n)
    {
        fac = new int[n + 1];
        for (int i = 1; i <= n; i++) fac[i] = -1;
        fac[0] = 1;
    }

    public int Ncr(int n, int r)
    {
        if (n < r) return 0;
        if (r == 0 | n == r) return 1;
        long a = Fac(n);
        a = a * Pow(Fac(n - r), p - 2) % p;
        a = a * Pow(Fac(r), p - 2) % p;
        return (int)a;
    }

    public int Fac(int n)
    {
        if (fac[n] >= 0) return fac[n];
        long k = 1;
        for (long i = 1; i <= n; i++)
        {
            k = k * i % p;
        }
        fac[n] = (int)k;
        return fac[n];
    }

    public int Pow(int n, int m)
    {
        if (m == 0) return 1;
        if (m == 1) return n;
        long a = 1;
        if (m % 2 == 0)
        {
            a = Pow(n, m / 2);
            a = a * a % p;
        }
        else
        {
            a = Pow(n, m / 2);
            a = a * a % p;
            a = a * n % p;
        }
        return (int)a;
    }
}