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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma region include
#include <iostream>
#include <iomanip>
#include <stdio.h>

#include <sstream>
#include <algorithm>
#include <cmath>
#include <complex>

#include <string>
#include <cstring>
#include <vector>
#include <tuple>
#include <bitset>

#include <queue>
#include <complex>
#include <set>
#include <map>
#include <stack>
#include <list>

#include <fstream>
#include <random>
//#include <time.h>
#include <ctime>
#pragma endregion //#include
/////////
#define REP(i, x, n) for(int i = x; i < n; ++i)
#define rep(i,n) REP(i,0,n)
/////////
#pragma region typedef
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
#pragma endregion //typedef
////定数
const int INF = (int)1e9;
const LL MOD = (LL)1e9+7;
const LL LINF = (LL)1e18;
const double PI = acos(-1.0);
const double EPS = 1e-9;
/////////
using namespace::std;

/*
thx
http://arc059.contest.atcoder.jp/submissions/961467
*/
void solve(){
	int N,C;
	cin >> N >> C;
	vector<int> A(N),B(N);
	for(int i=0;i<N;++i){
		cin >> A[i];
	}
	for(int i=0;i<N;++i){
		cin >> B[i];
	}
	
	//ppow[x][c] := x^cのテーブル
	vector< vector<LL> > ppow(401,vector<LL>(401,0));
	for(int x=1;x<=400;++x){
		LL temp = 1;
		ppow[x][0] = temp;//x^0;
		for(int c=1;c<=400;++c){
			temp = (temp * x ) % MOD;
			ppow[x][c] = temp;
		}
	}
	//ppow[x][c]のxについての累積和をとる
	vector< vector<LL> > sumpow(401,vector<LL>(401,0));
	for(int c=0;c<=400;++c){
		for(int x=1;x<=400;++x){
			sumpow[x][c] = (sumpow[x-1][c] + ppow[x][c])%MOD;
		}
	}

	vector< vector<LL> > dp(N+1,vector<LL>(C+1,0));
	dp[0][0] = 1;
	
	for(int i=0;i<N;++i){
		for(int now=0;now<=C;++now){//now個までくばった
			//dp[i][now]
			const int saMAX = C-now;//配れる残り個数
			for(int sa=0; sa<=saMAX; ++sa){//これからsa個くばる
				dp[i+1][now+sa] = (dp[i+1][now+sa] + 
					dp[i][now] * (sumpow[B[i]][sa]-sumpow[A[i]-1][sa])
					)%MOD;
			}
		}
	}
	cout << dp[N][C] << endl;
}

#pragma region main
signed main(void){
	std::cin.tie(0);
	std::ios::sync_with_stdio(false);
	std::cout << std::fixed;//小数を10進数表示
	cout << setprecision(16);//小数点以下の桁数を指定//coutとcerrで別	
	
	solve();
}
#pragma endregion //main()