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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for (int i = a; i < b; i++)
#define rep(i, n) REP(i, 0, n)
#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef unsigned long long ull;

const ll DIVIDER = 1000000007;

int main(){
    iostream::sync_with_stdio(false);
    ll N, K;
    cin >> N >> K;
    int A[2 * N];
    rep(i, N){
        cin >> A[i];
        A[N+i]=A[i];
    }
    ll scount1 = 0, scount2 = 0, lcount = 0;
    rep(i, N-1){
        REP(j, i+1, N){
            if (A[i] > A[j]){
                scount1++;
            }
        }
    }
    rep(i, 2*N-1){
        REP(j, i+1, 2*N){
            if (A[i] > A[j]){
                scount2++;
            }
        }
    }
    ll bcount = scount2-2*scount1;
    lcount += (scount1*K) % DIVIDER;
    REP(i, 1, K){
        lcount += bcount*i;
        if (lcount >= DIVIDER){
            lcount %= DIVIDER;
        }
    }
    cout << lcount << endl;
}