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;


#define li			long long int
#define rep(i,to)	for(li i=0;i<((li)(to));i++)
#define repp(i,start,to)	for(li i=(li)(start);i<((li)(to));i++)
#define pb			push_back
#define sz(v)		((li)(v).size())
#define bgn(v)		((v).begin())
#define eend(v)		((v).end())
#define allof(v)	(v).begin(), (v).end()
#define dodp(v,n)		memset(v,(li)n,sizeof(v))
#define bit(n)		(1ll<<(li)(n))
#define mp(a,b)		make_pair(a,b)
#define rin	rep(i,n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1


#define idp if(DBGP)
#define F first
#define S second
#define p2(a,b)		idp cout<<a<<"\t"<<b<<endl
#define p3(a,b,c)		idp cout<<a<<"\t"<<b<<"\t"<<c<<endl
#define p4(a,b,c,d)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<endl
#define p5(a,b,c,d,e)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<endl
#define p6(a,b,c,d,e,f)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<endl
#define p7(a,b,c,d,e,f,g)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<endl
#define p8(a,b,c,d,e,f,g,h)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<endl
#define p9(a,b,c,d,e,f,g,h,i)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<endl
#define p10(a,b,c,d,e,f,g,h,i,j)		idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<"\t"<<j<<endl
#define foreach(it,v)	for(__typeof((v).begin()) it=(v).begin(); it!=(v).end(); ++it)
#define p2p(x)		idp p2((x).F, (x).S)
#define dump(x,n)	idp{rep(i,n){cout<<x[i]<<" ";}puts("");}
#define dump2(x,n)	idp{rep(i,n){cout<<"["<<x[i].F<<" , "<<x[i].S<<"] ";}puts("");}
#define dumpi(x)	idp{foreach(it, x){cout<<(*it)<<" ";}puts("");}
#define dumpi2(x)	idp{foreach(it, x){cout<<"["<<(it)->F<<" , "<<(it)->S<<"] ";}puts("");}

#define read2d(a,w,h)	rep(i,h)rep(j,w)cin>>a[i][j]
#define dump2d(a,w,h)	rep(i,h){rep(j,w)cout<<a[i][j]<<" ";puts("");}

typedef pair<li, li> PI;


// dp[i][j]: j cookies before i
li dp[404][404];
li n, c;
li a[404];
li b[404];
li njo_sum_table[404][404];

// calc x^n%mod
inline li njo(li x, li nn, li mod) {
	//n=(n%(mod-1));
	if (nn < 1)return 1 % mod;
	if (nn == 1)return x % mod;
	li half = njo(x, nn / 2, mod);
	li double_half = (half * half) % mod;
	return (nn % 2 > 0 ? double_half * x % mod : double_half);
}

inline li njo2(li aa, li bb, li n, li mod) {
	//li res = 0;
	//repp(i, aa, bb + 1)res = (res + njo(i, n, mod)) % mod;
	//return res;
	return (njo_sum_table[n][bb] - njo_sum_table[n][aa - 1] + MOD) % MOD;
}

// njo_sum_table[n][x] contains {1^n+2^n+...+x^n}
inline void make_table() {
	rep(i, 404) {
		njo_sum_table[i][0] = 0;
		repp(j, 1, 404) {
			njo_sum_table[i][j] = (njo_sum_table[i][j - 1] + njo(j, i, MOD)) % MOD;
		}
	}
}

int main() {
	cin >> n >> c;
	rin{cin >> a[i];}
	rin{cin >> b[i];}

	make_table();
	//dump2d(njo_sum_table, c + 1, n + 1);

	rep(i, c + 1) {
		dp[0][i] = njo2(a[0], b[0], i, MOD);
	}
	repp(i, 1, n) {
		dp[i][0] = (dp[i - 1][0] * njo2(a[i], b[i], 0, MOD)) % MOD;
		repp(j, 1, c + 1) {
			dp[i][j] = 0;
			rep(k, j + 1) {
				dp[i][j] += (dp[i - 1][k] * njo2(a[i], b[i], j - k, MOD)) % MOD;
			}
			dp[i][j] %= MOD;
		}
	}
	cout << dp[n - 1][c] << endl;
	//dump2d(dp, c + 1, n);

	return 0;
}