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 repr(i,from,to) for(int (i)=(from);(i)<(to);(i)++)
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
// const bool debug=true;
const bool debug=false;
#define DEBUG if(debug==true)
#define vprint(x) for(auto a:(x)) cout << x << endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll MOD = 1000000007;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) { a = b; return true;}
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) { a = b; return true;}
    return false;
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while(n > 0) {
        if(n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n,c;cin>>n>>c;
    vector<ll> a(n), b(n);
    rep(i,n){cin>>a[i];}
    rep(i,n){cin>>b[i];}
    ll lim = 401;
    vector<vector<ll>> dp(n+1, vector<ll>(c+1, 0));
    vector<vector<ll>> sum(n+1,vector<ll>(c+1, 0));
    rep(i,n){
        rep(j,c+1){
            repr(k,a[i], b[i]+1){
                sum[i][j] += modpow(k, j, MOD);
                sum[i][j] %= MOD;
            }
        }
    }
    dp[0][0]=1;
    rep(i,n){
        rep(j,c+1){
            for(int k=0;j+k<=c;k++){
                dp[i+1][j+k] += dp[i][j]*sum[i][k];
                dp[i+1][j+k] %= MOD;
            }
        }
    }
    cout << dp[n][c] << endl;
    return 0;
}