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 <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <string>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <numeric>
#include <deque>
#include <cstring>
#include <iterator>
#include <map>
#include <cstdlib>
#include <unordered_map>
#include <unordered_set>
#include <iomanip>
#include <complex>
#include <bitset>
#include <chrono>
#include <random>
#include <assert.h>

using namespace std;

#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define eb emplace_back
#define el '\n'
typedef  long long  ll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef vector<ll> vll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<pll> vpll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<bool> vb;


const int mod=1000000007;
//const int mod=998244353;


long long binpow(long long a, long long b, long long m) {
    a %= m;
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}


vll  mult(vll & a, vll &b){
    int n=(int)a.size();
    vll c(n);
    for(int i=0; i<n; i++){
        for(int j=0; j+i<n; j++){
            c[i+j]+=a[i]*b[j];
            c[i+j]%=mod;
        }
    }
    return c;
}



int main()
{   ios_base::sync_with_stdio(false);cin.tie(NULL);
    ll t,n,m,k,q,x,a,b,c,d,y,l,r,z;
    cin >> n >> c;
    vll A(n), B(n);
    for(int i=0; i<n; i++){
        cin >> A[i];
    }
    for(int i=0; i<n; i++){
        cin >> B[i];
    }
    vvll v(n,vll(c+1,0));
    for(int i=0; i<n; i++){
        v[i][0]=B[i]-A[i]+1;
        for(int j=1; j<=c; j++){
            for(int k=A[i]; k<=B[i]; k++){
                v[i][j]+=binpow(k,j,mod);
            }
            v[i][j]%=mod;
        }
    }
//    print(v);
    for(int i=1; i<n; i++){
        v[i]=mult(v[i-1],v[i]);
    }
    cout << v[n-1][c] << el;
    return 0;
}