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++11 (GCC 4.8.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define all(c) ((c).begin()), ((c).end())
#define dump(c) cerr << "> " << #c << " = " << (c) << endl;
#define iter(c) __typeof((c).begin())
#define tr(i, c) for (iter(c) i = (c).begin(); i != (c).end(); i++)
#define REP(i, a, b) for (int i = a; i < (int)(b); i++)
#define rep(i, n) REP(i, 0, n)
#define mp make_pair
#define fst first
#define snd second
#define pb push_back
#define debug(fmt, ...) \
        fprintf( stderr, \
                  fmt "\n", \
                  ##__VA_ARGS__ \
        )

typedef unsigned int uint;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1 << 29;
const double EPS = 1e-10;

double zero(double d) {
    return d < EPS ? 0.0 : d;
}

template<class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );

template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
    return os << '(' << p.first << ',' << p.second << ')';
}

template<typename T>
ostream &operator<<(ostream &os, const vector<T> &a) {
    os << '[';
    rep(i, a.size()) os << (i ? " " : "") << a[i];
    return os << ']';
}

string toString(int i) {
    stringstream ss;
    ss << i;
    return ss.str();
}

const int MOD = 1000000007;

// a^k
ll fpow(ll a, ll k, int M) {
    ll res = 1ll;
    ll x = a;
    while (k != 0) {
        if ((k & 1) == 1)
            res = (res * x) % M;
        x = (x * x) % M;
        k >>= 1;
    }
    return res;
}

struct prepare {
    prepare() {
        cout.setf(ios::fixed, ios::floatfield);
        cout.precision(8);
        ios_base::sync_with_stdio(false);
    }
} _prepare;

int main() {
    int N, C;
    cin >> N >> C;
    vi A(N), B(N);
    rep(i, N)cin >> A[i];
    rep(i, N)cin >> B[i];

    // partial

    vvll dp(N+1, vll(C+1, 0));
    dp[0][0] = 1;
    rep(i, N) {
        rep(c, C+1) {
            dp[i+1][c] = dp[i][c];
            if(c > 0) {
                dp[i+1][c] += (dp[i+1][c-1] * A[i]) % MOD;
                dp[i+1][c] %= MOD;
            }
        }
    }

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

    return 0;
}