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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long   // <-----!!!!!!!!!!!!!!!!!!!

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n)-1;i>=0;i--)
#define rrep2(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define printV(v) for(auto&& x : v){cout << x << " ";} cout << endl
#define printVV(vv) for(auto&& v : vv){for(auto&& x : v){cout << x << " ";}cout << endl;}
#define printP(p) cout << p.first << " " << p.second << endl
#define printVP(vp) for(auto&& p : vp) printP(p);

typedef long long ll;
typedef pair<int, int> Pii;
typedef tuple<int, int, int> TUPLE;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<Pii> vp;
const int inf = 1e9;
const int mod = 1e9 + 7;
typedef vector<vector<int>> Graph;


signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    // 部分点解法 O(N * C^2)

    int N, C;
    cin >> N >> C;
    vi a(N + 1);
    rep(i, N) cin >> a[i + 1];
    vi b(N + 1);
    rep(i, N) cin >> b[i + 1];

    vvi dp(N + 1, vi(C + 1));
    dp[0][0] = 1;
    rep(i, N) {
        rep(j, C + 1) {
            // dp[i + 1][j] = sum{ dp[i][j - k] * x_(i+1) ^ k }; (k = 0..j)
            int t = 1;  // x_(i+1) ^ k の値
            rep(k, j + 1) {
                // dp[i + 1][j] += dp[i][j - k] * pow(a[i + 1], k, mod);
                dp[i + 1][j] += dp[i][j - k] * t;
                (t *= a[i + 1]) %= mod;
            }
        }
    }

    cout << dp[N][C] << endl;

}