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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int MOD = 1000000007;
//const int MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const double pi = acos(-1.0);
const double EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

template<int mod>
struct Mod_Int{
    ll x;

    Mod_Int() : x(0) {}

    Mod_Int(ll y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    Mod_Int &operator += (const Mod_Int &p){
        x = (x + p.x) % mod;
        return *this;
    }

    Mod_Int &operator -= (const Mod_Int &p){
        x = (x + mod - p.x) % mod;
        return *this;
    }

    Mod_Int &operator *= (const Mod_Int &p){
        x = (x * p.x) % mod;
        return *this;
    }

    Mod_Int &operator /= (const Mod_Int &p){
        *this *= p.inverse();
        return *this;
    }

    Mod_Int &operator ++ () {return *this += Mod_Int(1);}

    Mod_Int operator ++ (int){
        Mod_Int tmp = *this;
        ++*this;
        return tmp;
    }

    Mod_Int &operator -- () {return *this -= Mod_Int(1);}

    Mod_Int operator -- (int){
        Mod_Int tmp = *this;
        --*this;
        return tmp;
    }

    Mod_Int operator - () const {return Mod_Int(-x);}

    Mod_Int operator + (const Mod_Int &p) const {return Mod_Int(*this) += p;}

    Mod_Int operator - (const Mod_Int &p) const {return Mod_Int(*this) -= p;}

    Mod_Int operator * (const Mod_Int &p) const {return Mod_Int(*this) *= p;}

    Mod_Int operator / (const Mod_Int &p) const {return Mod_Int(*this) /= p;}

    bool operator == (const Mod_Int &p) const {return x == p.x;}

    bool operator != (const Mod_Int &p) const {return x != p.x;}

    Mod_Int inverse() const {
        assert(*this != Mod_Int(0));
        return pow(mod-2);
    }

    Mod_Int pow(ll k) const{
        Mod_Int now = *this, ret = 1;
        while(k){
            if(k&1) ret *= now;
            now *= now, k >>= 1;
        }
        return ret;
    }

    friend ostream &operator << (ostream &os, const Mod_Int &p){
        return os << p.x;
    }

    friend istream &operator >> (istream &is, Mod_Int &p){
        ll a;
        is >> a;
        p = Mod_Int<mod>(a);
        return is;
    }
};

using mint = Mod_Int<MOD>;

int main(){
    int N, C;
    cin >> N >> C;
    int A[N], B[N];
    rep(i, N) cin >> A[i];
    rep(i, N) cin >> B[i];
    mint sum[401][401];
    rep(i, 401){
        sum[i][0] = 0;
        rep(j, 400) sum[i][j+1] += sum[i][j]+mint(j+1).pow(i);
    }
    mint dp[N+1][C+1];
    fill(dp[0], dp[N+1], 0);
    dp[0][0] = 1;
    rep(i, N){
        rep(j, C+1){
            rep(k, C-j+1){
                dp[i+1][j+k] += dp[i][j]*(sum[k][B[i]]-sum[k][A[i]-1]);
            }
        }
    }
    cout << dp[N][C] << endl;
}