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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, double> Pid;
typedef pair<double, int> Pdi;
const double PI = 3.1415926535897932;   // acos(-1)
const double EPS = 1e-15;
const int INF = 1001001001;
const ll mod = 1e+9 + 7;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define chadd(x, y) x = (x + y) % mod

ll dp[405][405];
ll sum[405][405];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, c;
    cin >> n >> c;
    vector<int> a(n), b(n);
    for(int i = 0; i < n; ++i)  cin >> a[i];
    for(int i = 0; i < n; ++i)  cin >> b[i];

    int maxe = max(*max_element(a.begin(), a.end()), *max_element(b.begin(), b.end()));
    for(int i = 0; i <= maxe; ++i)  sum[i][0] = 1;
    for(int i = 1; i <= maxe; ++i){
        for(int j = 0; j < c; ++j){
            sum[i][j+1] = sum[i][j] * i % mod;
        }
    }
    for(int i = 0; i < maxe; ++i){
        for(int j = 0; j <= c; ++j){
            chadd(sum[i+1][j], sum[i][j]);
        }
    }

    dp[0][0] = 1;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j <= c; ++j){
            for(int k = 0; k <= j; ++k){
                // a_i^k + (a_i + 1)^k + ... + b[i]^k
                ll mul = (sum[b[i]][k] - sum[a[i]-1][k]) % mod;
                if(mul < 0) mul += mod;
                ll add = dp[i][j-k] * mul % mod;
                chadd(dp[i+1][j], add);
            }
        }
    }

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