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 <iomanip>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
using namespace std;

template <int64_t M>
class mint {
public:
        mint() :v(-1) {}
        mint(int n) : v(n % M) {}

        int64_t get() const { return v; }

        mint &operator+=(const mint &r) {
                v = (v + r.v) % M;
                return *this;
        }
        mint &operator-=(const mint &r) {
                v = (v - r.v + M) % M;
                return *this;
        }
        mint &operator*=(const mint &r) {
                v = (v * r.v) % M;
                return *this;
        }
        mint &operator/=(const mint &r) {
                return (*this) *= r.inv();
        }

        mint pow(int r) const {
                mint k = *this, ret = 1;
                while (r > 0) {
                        if (r % 2 != 0) ret *= k;
                        r /= 2;
                        k *= k;
                }
                return ret;
        }

        mint inv() const {
                return pow(M - 2);
        }

private:
        int64_t v;
};

template <int64_t M>
inline mint<M> operator+(mint<M> l, const mint<M> &r) {
        return l += r;
}

template <int64_t M>
inline mint<M> operator-(mint<M> l, const mint<M> &r) {
        return l -= r;
}

template <int64_t M>
inline mint<M> operator*(mint<M> l, const mint<M> &r) {
        return l *= r;
}

template <int64_t M>
inline mint<M> operator/(mint<M> l, const mint<M> &r) {
        return l /= r;
}

template <int64_t M>
inline ostream &operator<<(ostream &os, const mint<M> &r) {
        return os << r.get();
}

typedef mint<1000000007> MINT;

int n, c;
vector<int> as, bs;

MINT powt[401][401];
MINT pt[401][401];

void calc()
{
    for (int i = 0; i < n; i++) {

        for (int j = 0; j <= c; j++) {
            MINT sum = 0;

            for (int k = as[i]; k <= bs[i]; k++) sum += powt[k][j];

            pt[i][j] = sum;
        }
    }
}

MINT tbl[401][401];
bool flg[401][401];

MINT solve(int cur, int rest)
{
    if (cur == n) return rest == 0 ? 1 : 0;

    MINT &ret = tbl[cur][rest];
    if (flg[cur][rest]) return ret;

    flg[cur][rest] = true;
    ret = MINT(0);

    MINT cc = 1;
    for (int i = 0; i <= rest; i++, cc *= as[cur]) {

        ret += solve(cur + 1, rest - i) * pt[cur][i];
    }

    return ret;
}

int main()
{
    cin >> n >> c;

    as.resize(n);
    bs.resize(n);

    for (auto &a: as) cin >> a;
    for (auto &b: bs) cin >> b;

    for (int i = 0; i <= 400; i++) {
        powt[i][0] = MINT(1);
        for (int j = 1; j <= 400; j++) {
            powt[i][j] = powt[i][j-1] * MINT(i);
        }
    }
    calc();

    memset(&flg, 0, sizeof(flg));
    cout << solve(0, c) << endl;

    return 0;
}