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.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
static long factMax = 2000;//nCkにおけるnの最大値
static long[] factorials = new long[factMax+1];//i!(mod p)をO(n)で先にメモ。
static long[] factorialRs = new long[factMax+1];//i!^-1(mod p), pは素数でO(n)。

	static void Main()
	{
    long mod = 1000000007;
    long answer = 0;
    factorials[0] = 1;
    factorialRs[factMax] = DivideModFactorial(factMax,mod);
    for(long i = 1; i <= factMax; i++)
    {
      factorials[i] = (factorials[i-1]*i)%mod;//i!(mod)
      factorialRs[factMax-i] = (factorialRs[factMax+1-i]*(factMax+1-i))%mod;//逆元も先にメモ
    }
  
		string[] input = Console.ReadLine().Split(' ');
		int n = int.Parse(input[0]);
		int c = int.Parse(input[1]);
    long[,] dp = new long[n+1,c+1];//n個見てc個配った
		long[] numsA = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
    long[] numsB = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
    long[,] pows = new long[401,401];//aのb乗
    for(long i = 1; i <= 400; i++)
    {
      long powMemo = 1;
      for(int j = 0; j <= 400; j++)
      {
        pows[(int)i,j] = powMemo;
        powMemo *= i;
        powMemo %= mod;
      }
    }

    long[,] xPows = new long[401,401];//1^a +2^a +b^a
    for(int i = 0; i <= 400; i++)
    {
      long powMemo = 0;
      for(int j = 0; j <= 400; j++)
      {
        xPows[i,j] = powMemo;
        powMemo += DivideMod(j+1,i,mod);
        powMemo %= mod; 
      }
    }

    for(int i = 0; i <= c; i++) dp[0,0] = 1;
    
    for(int i = 1; i <= n; i++)
    {
      for(int j = 0; j <= c; j++)
      {
        for(int k = 0; k <= 400; k++)
        {
          if(j-k >= 0) dp[i,j] += dp[i-1,j-k]*(xPows[k,numsB[i-1]]-xPows[k,numsA[i-1]-1]);
          dp[i,j] %= mod;
        }
        //Console.WriteLine(i+" "+j+" "+dp[i,j]);
      }
    }

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

  static long MathPow(long x, long a)//戻り値はx^a。O(loga)
  {
    long num = 2;
    long answer = 1;
    long check = a;
    long memo = x;
  
    while(check > 0)
    {
      if(check % num == num / 2)
      {
        check -= num / 2;
        answer *= memo;
      }
    memo *= memo;
    num *= 2;
    }
    return answer;
  }

  static long DivideMod(long x, long a, long p)//戻り値はx^a(mod p)。O(loga)
  {
    long num = 2;
    long answer = 1;
    long check = a;
    long memo = x%p;
    
    while(check > 0)
    {
      if(check % num == num / 2)
      {
        check -= num / 2;
        answer *= memo;
        answer %= p;
      }
    memo *= memo;
    memo %= p;
    num *= 2;
    }
    return answer;
  }

  static long DivideModReverse(long x, long p)//戻り値はx^-1(mod p), pは素数。O(logp)
  {
    long answer = DivideMod(x, p-2, p);
    return answer;
  }

  static long ModFactorial(long x, long p)//戻り値はx!(mod p)
  {
    long answer = 1;
    for(int i = 2; i <= x; i++)
    {
      answer *= i;
      answer %= p;
    }
    return answer;
  }

  static long DivideModFactorial(long x, long p)//戻り値はx!^-1(mod p), pは素数
  {
    long answer = 1;
    for(long i = x; i >= 2; i--)
    {
      answer *= DivideModReverse(i, p);
      answer %= p;
    }
    return answer;
  }
  
}