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<algorithm>
#include<iostream>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<bitset>
#include<cctype>
#include<stack>
#include<queue>
#include<deque>
#include<cmath>
#include<map>
#include<set>
#define LL long long
#define ULL unsigned  long long
#define maxn (LL)1e5
#define INF 0x3f3f3f3f
#define inf 0x7fffffff
#define PI  acos(-1.0)
#define pb push_back
#define re register LL
const double eps = 0.0000001;
using namespace std;
typedef pair<LL,LL> pii;
inline LL sgn(double x)
{
    return (x > eps) - (x < -eps);
}
const LL N=1e6+2;
const LL mod=1e9+7;
LL f[2010];
int main()
{
   LL n,k;
   cin>>n>>k;
   for(LL i=1;i<=n;i++)
   {
       cin>>f[i];
   }
   LL num1=0;
   for(LL i=1;i<=n;i++)
   {
       for(LL j=1;j<=n;j++)
       {
           if(f[i]>f[j]) num1++;
       }
   }
   LL num=0;
   for(LL i=1;i<n;i++)
   {
       for(LL j=i+1;j<=n;j++)
       if(f[j]<f[i]&&j>i) num++;
   }
   cout<<(num*k+(k*(k-1)/2)*num1)%mod;
}