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: Accepted)

#include <bits/stdc++.h>
using namespace std;
template<class T> ostream& operator<<(ostream &os, vector<T> V) { os << "[ "; 
	for(auto v:V)os << v << " "; return os << "]";
}
template<class T> ostream& operator<<(ostream &os, set<T> S){ os << "{ "; 
	for(auto s:S) os<<s<<" ";  return os<<"}"; 
}
template<class L, class R> ostream& operator<<(ostream &os, pair<L,R> P) {
	return os<<"("<<P.first<<","<< P.second << ")";
}
template<class L, class R> ostream& operator<<(ostream &os, map<L,R> M) {
	os<<"{ ";for(auto m:M)os<<"("<<m.F<<":"<<m.S<<")";
	return os<<"}";
}
#define cerr cout
#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...) 1
#endif
#define ll long long
#define ld long double
#define vll vector<ll>
#define pll pair<ll,ll>
#define vpll vector<pll>
#define I insert 
#define pb push_back
#define F first
#define S second
#define endl "\n"
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define For(i,n) for(int i=0;i<(int)n;++i)
#define Rev(i,n) for(int i=(int)n-1;i>=0;--i)
#define Rep(i,n) for(int i=1;i<=(int)n;++i)
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef pair<pii,int> ppi;
typedef vector<pii> vpii;
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int N = 405;
const int mod = 1e9+7;

inline int mul(int a, int b){
	return a*1ll*b%mod;
}
inline int add(int a ,int b){
	a += b;
	if(a >= mod) a -= mod;
	return a;
}
int power(int x , int y){
	int res = 1;
	while(y){
		if(y&1)
			res = mul(res , x);
		x = mul(x , x);
		y >>= 1;
	}
	return res;
}
int A[N] , B[N];

int dp[N][N]; //number of ways to divide j candies to first i students

int main(){
	fio; cout<<fixed<<setprecision(25);	
	int n , c; cin >> n >> c;

	for(int i = 1;i <= n ; ++i){
		cin >> A[i];
	}
	for(int i =1 ;i <= n ; ++i){
		cin >> B[i];
	}
	dp[0][0] = 1;
	for(int i =1 ;i <= n; ++i){
		for(int j = 0;j <= c ; ++j){
			//number of candies to curr student;
			int curr = 0;
			for(int k = A[i] ; k <= B[i] ; ++k)
				curr = add(curr , power(k , j));

			for(int k = j; k <= c ; ++k){
				dp[i][k] = add(dp[i][k] , mul(dp[i-1][k-j] , curr));
			}
		}
	}
	cout << dp[n][c] << endl;

	return 0;
}