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: Accepted)

using System;
using System.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Collections.Generic;
using System.Diagnostics;
using System.Numerics;
using Enu = System.Linq.Enumerable;

public class Program
{
    static readonly int Mod = (int)1e9 + 7;
    static void Add(ref long a, long b) { if ((a += b) >= Mod) a -= Mod; }


    public void Solve()
    {
        int N = Reader.Int(), C = Reader.Int();
        var A = Reader.IntArray(N);
        var B = Reader.IntArray(N);

        int Bmax = B.Max();
        var powSum = new long[Bmax + 1, C + 1];
        for (int x = 1; x <= Bmax; x++)
            for (long n = 0, p = 1; n <= C; n++, p = p * x % Mod)
                powSum[x, n] = (powSum[x - 1, n] + p) % Mod;

        var dp = new long[C + 1];
        dp[0] = 1;
        for (int i = 0; i < N; i++)
        {
            var nextdp = new long[C + 1];
            for (int sum = 0; sum <= C; sum++)
                if (dp[sum] > 0)
                    for (int add = 0; sum + add <= C; add++)
                    {
                        long v = Mod + powSum[B[i], add] - powSum[A[i] - 1, add];
                        Add(ref nextdp[sum + add], v % Mod * dp[sum] % Mod);
                    }
            dp = nextdp;
        }

        Console.WriteLine(dp[C] % Mod);
    }

}


class Entry { static void Main() { new Program().Solve(); } }
class Reader
{
    static TextReader reader = Console.In;
    static readonly char[] separator = { ' ' };
    static readonly StringSplitOptions op = StringSplitOptions.RemoveEmptyEntries;
    static string[] A = new string[0];
    static int i;
    static void Init() { A = new string[0]; }
    public static void Set(TextReader r) { reader = r; Init(); }
    public static void Set(string file) { reader = new StreamReader(file); Init(); }
    public static bool HasNext() { return CheckNext(); }
    public static string String() { return Next(); }
    public static int Int() { return int.Parse(Next()); }
    public static long Long() { return long.Parse(Next()); }
    public static double Double() { return double.Parse(Next()); }
    public static int[] IntLine() { return Array.ConvertAll(Split(Line()), int.Parse); }
    public static int[] IntArray(int N) { return Range(N, Int); }
    public static int[][] IntTable(int H) { return Range(H, IntLine); }
    public static string[] StringArray(int N) { return Range(N, Next); }
    public static string[][] StringTable(int N) { return Range(N, () => Split(Line())); }
    public static string Line() { return reader.ReadLine().Trim(); }
    public static T[] Range<T>(int N, Func<T> f) { var r = new T[N]; for (int i = 0; i < N; r[i++] = f()) ; return r; }
    static string[] Split(string s) { return s.Split(separator, op); }
    static string Next() { CheckNext(); return A[i++]; }
    static bool CheckNext()
    {
        if (i < A.Length) return true;
        string line = reader.ReadLine();
        if (line == null) return false;
        if (line == "") return CheckNext();
        A = Split(line);
        i = 0;
        return true;
    }
}