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)

/* Generated by powerful Codeforces Tool
 * You can download the binary file in here https://github.com/xalanq/cf-tool (Windows, macOS, Linux)
 * Author: alireza_kaviani
 * Time: 2020-05-14 11:40:42
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;

#define all(x)                      (x).begin(),(x).end()
#define Sort(x)                     sort(all((x)))
#define X                           first
#define Y                           second
#define sep                         ' '
#define endl                        '\n'
#define SZ(x)                       ll(x.size())

ll poww(ll a, ll b, ll md) {
    return (!b ? 1 : (b & 1 ? a * poww(a * a % md, b / 2, md) % md : poww(a * a % md, b / 2, md) % md));
}

const ll MAXN = 400 + 10;
const ll LOG = 22;
const ll INF = 8e18;
const ll MOD = 1e9 + 7; // 998244353; // 1e9 + 9;

ll n , C , A[MAXN] , B[MAXN] , dp[MAXN][MAXN] , pw[MAXN][MAXN];

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

    for(int i = 1 ; i < MAXN ; i++){
        ll cur = 1;
        for(int j = 0 ; j < MAXN ; j++){
            pw[i][j] = (pw[i - 1][j] + cur) % MOD;
            cur = cur * i % MOD;
        }
    }

    cin >> n >> C;
    for(int i = 1 ; i <= n ; i++)   cin >> A[i];
    for(int i = 1 ; i <= n ; i++)   cin >> B[i];

    dp[0][0] = 1;
    for(int i = 1 ; i <= n ; i++){
        for(int j = 0 ; j <= C ; j++){
            for(int k = j ; k <= C ; k++){
                dp[i][k] = (dp[i][k] + dp[i - 1][j] * (pw[B[i]][k - j] - pw[A[i] - 1][k - j] + MOD)) % MOD;
            }
        }
    }
    cout << dp[n][C] << endl;

    return 0;
}
/*

*/