Problem:
Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, n) for(ll i=0; i<n; i++)
#define SORT(x) sort(x.begin(), x.end());
using namespace std;
typedef long long ll;

const ll MOD = 1000000007;

int main() {
    ll N, K;
    cin >> N >> K;
    vector<ll> A(N);
    rep(i, N) cin >> A[i];

    ll temp1 = 0;
    for(ll i=0; i<N; i++) {
        for(ll j=i+1; j<N; j++) {
            if(A[i] > A[j]) temp1++;
        }
    }
    temp1 %= MOD;

    ll temp2 = 0;
    for(ll i=0; i<N; i++) {
        for(ll j=0; j<N; j++) {
            if(A[i] > A[j]) temp2++;
        }
    }
    temp2 %= MOD;

    ll temp3 = (temp1 * K) % MOD;
    ll temp4 = (temp2 * (K-1)) % MOD;
    ll res;
    if(K > 10000) {
        int k;
        ll temp5 = 0;
        for(k=0; k<K; k+=10000) {
            temp5 += (temp4 * 10000) % MOD;
        }
        temp5 += (temp4 * (K-k)) % MOD;
        temp5 %= MOD;
        temp5 /= 2;
        temp5 %= MOD;
        res = (temp3 + temp5) % MOD;
    } else {
        temp4 *= (K-1);
        temp4 /= 2;
        temp4 %= MOD;
        res = (temp3 + temp4) % MOD; 
    }

    cout << res << endl;
    
    return 0;
}