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>
using namespace std;

#define _p(...) (void)printf(__VA_ARGS__)
#define forr(x,arr) for(auto&& x:arr)
#define _overload3(_1,_2,_3,name,...) name
#define _rep2(i,n) _rep3(i,0,n)
#define _rep3(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,_rep3,_rep2,)(__VA_ARGS__)
#define _rrep2(i,n) _rrep3(i,0,n)
#define _rrep3(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define rrep(...) _overload3(__VA_ARGS__,_rrep3,_rrep2,)(__VA_ARGS__)
#define ALL(x) (x).begin(), (x).end()
#define BIT(n) (1LL<<(n))
#define SZ(x) ((int)(x).size())
#define fst first
#define snd second
typedef vector<int> vi;typedef vector<vi> vvi;typedef pair<int,int> pii;typedef vector<pii> vpii;
typedef long long ll;

typedef vector<ll> vl;typedef vector<vl> vvl;

const long long mod = 1000000007;

long long powMod(long long x, long long y) {
  long long r = 1, a = x % mod;
  while (y) {
    if (y & 1) r = (r * a) % mod;
    a = a * a % mod;
    y /= 2;
  }
  return r;
}

void Main() {
  int N, K;
  scanf("%d%d", &N, &K);
  vi A(N), B(N);
  rep(i, N) scanf("%d", &A[i]);
  rep(i, N) scanf("%d", &B[i]);

  vvl dp(N+1, vl(K+1));
  dp[0][0] = 1;

  // ppow[i][j] : i^j
  vvl ppow(401, vl(K+1));
  rep(i, 401) {
    ppow[i][0] = 1;
    rep(j, K) {
      (ppow[i][j+1] = ppow[i][j] * i % mod) %= mod;
    }
  }

  // spow[i][j] : 0^j + 1^j + 2^j + ... + (i-1)^j
  vvl spow(401+1, vl(K+1));
  rep(i, 401) {
    rep(j, K+1) {
      spow[i+1][j] = (spow[i][j] + ppow[i][j]) % mod;
    }
  }

  rep(i, N) rep(j, K+1) {
    // dp[i][j] : i-1 番までに j 個
    int a = A[i];
    int b = B[i];
    rep(k, j+1) {
      // i番がk個
      // p : (1^k + 2^k + 3^k + ...b^k) - (1^k + 2^k + 3^k + ... + (a-1)^k) = a^k + (a+1)^k + ... + b^k
      ll p = (spow[b+1][k] - spow[a][k] + mod) % mod;
      (dp[i+1][j] += dp[i][j-k] * p) %= mod;
    }
  }

  ll ans = dp[N][K];

  _p("%lld\n", ans);
}
int main() { cin.tie(0); ios::sync_with_stdio(false); Main(); return 0; }