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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define md 1000000007

int main()
{
    ll n, i,  j, k, m, p=0, c, pp, h=0, hh;
    cin>>n>>k;

    ll a[n], b[n], s[n], ss[n];
    for(i=0; i<n; i++){
        cin>>a[i];
        b[i]=a[i];
    }
    sort(b, b+n);
    s[b[0]]=0;
    for(i=1; i<n; i++){
        if(b[i]!=b[i-1]){
           // g[b[i-1]]=n-i;
            s[b[i]]=i;
        }
    }
    for(i=0; i<n; i++){
        p=0;
        for(j=i-1; j>=0; j--){
            if(a[j]<a[i]) p++;
        }
        ss[i]=p;
    }
    //g[b[n-1]]=0;
    p=0;
    for(i=0; i<n; i++){
        p=((k*s[a[i]])-ss[i])%md;
        pp=s[a[i]]-ss[i];
        hh=((k*(2*pp + (k-1)*s[a[i]]))%md)/2;
        h=(h+hh)%md;
    }
    cout<<h<<endl;
    return 0;
}