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)

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<tuple>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define Per(i,sta,n) for(int i=n-1;i>=sta;i--)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef long double ld;
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ll, ll> LP;

int n,a[500],b[500];
ll c;
ll x[500][500],dp[500][500];

struct mod_fac{
    private:
        vector<ll> f,i_f;
        int n;ll MOD;
    public:
        void init(int n,ll m){
            MOD=m;
            f.resize(n+1,0);
            i_f.resize(n+1,0);
            f[0]=1;
            rep(i,n){
                f[i+1]=f[i]*(i+1)%MOD;
            }
            i_f[n]=power(f[n],MOD-2);
            per(i,n){
                i_f[i]=(i_f[i+1]*(i+1))%MOD;
            }
        }
        ll power(ll a ,ll x){
            if (x==0) return 1;
            if (x==1) return a%MOD;
            if (x==2) return (a*a)%MOD;
            if (x%2==1) return (a*power(power(a,(x-1)/2),2))%MOD;
            return (power(power(a,x/2),2))%MOD;
        }
        ll fac(int x){
            return f[x];
        }
        ll invfac(int x){
            return i_f[x];
        }
        ll comb(int m,int k){
            if (m<0 or k<0) return 0;
            if (m<k) return 0;
            return (((f[m]*i_f[k])%MOD)*i_f[m-k])%MOD;
        }
};

void solve(){
    cin >> n >> c;
    rep(i,n+1){
        rep(k,c+1){
            dp[i][c]=0;
            x[i][c]=0;
        }
    }
    mod_fac m;m.init(1,mod);
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];
    rep(i,n){
        rep(k,c+1){
            Rep(j,a[i],b[i]+1) {
                x[i][k]+=m.power(j,k);
                x[i][k]%=mod;
            }
            //cout << i << " " << k << " " << x[i][k] << endl;
        }
    }

    dp[0][0]=1;
    rep(i,n){
        rep(k,c+1){
            rep(l,k+1){
                dp[i+1][k]+=dp[i][k-l]*x[i][l];
                dp[i+1][k]%=mod;
            }
        }
    }

    cout << dp[n][c] << endl;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    solve();
}