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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#define ll long long
#define rep(i,l,r,d) for(ll i=(l);i<(r);i+=(d))
ll A[2001], lowp[2001], lowa[2001];
int main(void){
    ll N, K, MOD=1000000007, k;
    scanf("%lld %lld", &N, &K);
    rep(i, 0, N, 1) scanf("%lld", A+i);
    //A[i]より後ろにいくつ小さいのがあるか
    rep(i, 0, N, 1) rep(j, i+1, N, 1) if(A[i]>A[j])lowp[i]++;
    //全体として小さいのがいくつあるか
    rep(i, 0, N, 1) rep(j, 0, N, 1) if(A[i]>A[j]) lowa[i]++;
    //for(i=K-1, i>0, i--) k += i の公式
    if(K%2==0)k=((K+1)*(K/2) - K)%MOD;
    else k=(K*(K/2))%MOD;
    rep(i, 0, N, 1){
        A[2000] += (lowp[i]*K)%MOD + (lowa[i]*k)%MOD;
        A[2000] = A[2000]%MOD;
    }
    printf("%lld\n", A[2000]);
    return 0;
}