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)

#include <bits/stdc++.h>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < int , int >
#define F first
#define S second
#define int long long
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize ("Ofast")
using namespace std;
/// khodaya komak kon
/// ya navid navid
/// ye tec khaphan ke yadam bemone : age jayi didi soale masir hamiltoni mikhad rasasho bokon yal oilery;
/// age ye ja mikhasti to dp az ozv i k ta entekhab koni bejash ye log bezan (nominal) hamoon 2**k va o 1 ent kon

///\\\///\\\///\\\\\\\\\\\\\\\\\///\\\///\\\///

const int N=410,mod=1e9+7;

ll dp[N][N];

ll a[N];
ll b[N];
ll p[N][N];
int32_t main(){
    ll n,c;
    cin >> n >> c;
    for (int i=1;i<N;i++){
        p[i][0]=1;
        for (int j=1;j<N;j++){
            p[i][j]=p[i][j-1]*i%mod;
        }
    }
    for (int i=0;i<n;i++){
        cin >> a[i+1];
    }
    for (int i=0;i<n;i++){
        cin >> b[i+1];
    }
    dp[0][c]=1;
    for (int i=1;i<=n;i++){
        for (int j=0;j<=c;j++){
            for (int l=0;l<=c;l++){
                /// dp[i][j]
                if (j+l<=c){
                    ll z=p[a[i]][l];
                    dp[i][j]+=z*dp[i-1][j+l]%mod;
                    dp[i][j]%=mod;
                }
            }
           // cout << i << " " << j << " " << dp[i][j] << endl;
        }
    }
    cout << dp[n][0];


}