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: Time Limit Exceeded)

    //Daniel Grzegorzewski
    #include <bits/stdc++.h>
     
    #define MP make_pair
    #define PB push_back
    #define ST first
    #define ND second
    #define int long long
     
    using namespace std;
     
    typedef pair<int, int> PII;
    typedef vector<int> VI;
    typedef vector<PII> VII;
    typedef long long LL;
     
    void init_ios()
    {
         ios_base::sync_with_stdio(0);
         cin.tie(0);
    }
     
    const int MOD = (int)1e9 + 7;
    const int N = 810;
     
    int n, c, a[N], b[N], sil[N], odwr[N], ile[N], dp[N][N], sumpot[N][N];
     
    int fpow(int x, int y)
    {
    	if (y == 0)
    		return 1;
    	if (y%2 == 1)
    		return (x*fpow(x, y-1))%MOD;
    	int res = fpow(x, y/2);
    	return (res*res)%MOD;
    }
     
    int C(int x, int y)
    {
    	if (x < y || x < 0 || y < 0)
    		return 0;
    	int res = (sil[x]*odwr[y])%MOD;
    	return (res*odwr[x-y])%MOD;
    }
     
    signed main()
    {
        init_ios();
        cin >> n >> c;
        sil[0] = odwr[0] = 1;
        for (int i = 1; i < N-5; ++i) {
        	sil[i] = (i*sil[i-1])%MOD;
        	odwr[i] = fpow(sil[i], MOD-2);
        }
        for (int i = 1; i <= n; ++i)
        	cin >> a[i];
        for (int i = 1; i <= n; ++i)
        	cin >> b[i];
        for (int i = 1; i <= n; ++i)
        	for (int j = 0; j <= c; ++j) {
        		int tmp = 0;
        		for (int k = a[i]; k <= b[i]; ++k)
        			sumpot[i][j] = (sumpot[i][j]+fpow(k, j))%MOD;
        	}
        for (int i = 0; i <= c; ++i)
        	dp[n][i] = sumpot[n][i];
        for (int i = n-1; i >= 1; --i)
        	for (int j = 0; j <= c; ++j)
        		for (int k = 0; k <= j; ++k)
        			dp[i][j] = (dp[i][j] + sumpot[i][k]*dp[i+1][j-k])%MOD;
        cout<<dp[1][c]<<"\n";
    }