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>
#include<iostream>
#include<cmath>
#define rep(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
using namespace std;
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const double eps = 0.0000001;
const int N=2200;
const int M=1e5+10;
const int mod=1e9+7;
void io(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
}
typedef pair<ll,ll> pii;
inline ll sgn(double x)
{
    return (x > eps) - (x < -eps);
}
int aa[2010];
ll p(ll a,ll b,ll mod)
{
    ll ans=0;
    for(;b;b>>=1)
    {
        if(b&1)
        {
            ans=ans+a%mod;
        }
        a=a*2%mod;
    }
    return ans%mod;
}
int main()
{
   int n;
   ll k;
   cin>>n>>k;
   for(int i=1;i<=n;i++)
       cin>>aa[i];
   int sum=0;
   for(int i=1;i<=n;i++)
       for(int j=1;j<=n;j++)
           if(aa[i]>aa[j]) sum++;
   int summ=0;
   for(int i=1;i+1<=n;i++)
       for(int j=i+1;j<=n;j++)
       if(aa[j]<aa[i]) summ++;
    ll ans=(ll)p((ll)sum,k,mod);
    ans+=p(k*(k-1)/2,(ll)sum,mod)%mod;
    cout<<ans%mod<<endl;
}