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>
#define SORT(x) sort((x).begin(), (x).end())
#define ALL(x) (x).begin(), (x).end()
#define rep(i, n) for (ll i = 0; i < n; i++)
#define reps(i, m, n) for (ll i = m; i < n; i++)
#define repr(i, m, n) for (ll i = m; i >= n; i--)
#define de(x) cout << #x << "=" << x << endl;
template <class T> bool maxi(T &a, const T &b) {
    if (a < b) { a = b; return 1;} return 0;}
template <class T> bool mini(T &a, const T &b) {
    if (b < a) { a = b; return 1;} return 0;}
#define dame {cout<< "-1" <<"\n"; return;}
#define INF2 1000000000000000037
#define INF  1000000007
#define MOD  1000000007
using namespace std;
using ll = long long;
using ld = long double;
using P  = pair<ll,ll>;
//--GLOBAL---------------------------------
ll dp[410][410]; // i番までにj個与える

ll modpow(ll a, ll n=MOD-2, ll mod=MOD){ // a^n % MOD の計算
    ll res = 1;
    while (n > 0){
        if(n & 1) res = res * a % mod;
        a = a * a  % mod;
        n >>= 1;
    }
    return res;
}

ll powawa[410][410];
//--MAIN-----------------------------------
void Main(){
    ll N,C; cin>>N>>C;
    vector<ll> A(N);
    vector<ll> B(N);
    rep(i,N) cin>>A[i];
    rep(i,N) cin>>B[i];

    dp[0][0]=1;

    rep(i,401){
        powawa[i][0]=1;
        reps(j,1,401){
            powawa[i][j]=(powawa[i][j-1]*i)%MOD;
        }
    }

    // sumj=0->i( j^x )
    ll sum[410][410];
    rep(i,405){
        rep(j,405){
            sum[i][j]=powawa[i][j];
        }
    }
    rep(i,405){
        rep(j,405){
            (sum[i+1][j]+=sum[i][j])%=MOD;
        }
    }



    rep(i,N){
        //reps(j,A[i],B[i]+1){ // A->Bのとき
            rep(k,C+1){ // iまででk個
                rep(l,C-k+1){ // C個になるまで配る l個
                    ll kakeru=((sum[B[i]][l]-sum[A[i]-1][l])+MOD)%MOD;
                    (dp[i+1][k+l]+=dp[i][k]*kakeru)%=MOD;
                }
            }   
        //}
    }
    cout<< dp[N][C] <<"\n";
}
//--START----------------------------------
int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
}
//-----------------------------------------