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>
using namespace std;
#define ll  long long
#define ull unsigned long long
#define pii pair<int,int>
#define pb(x) push_back(x)
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define drep(i,a,b) for(int i=(a); i>=(b);i--)
 
const int maxn = 1e5 + 5;
const int maxm = 5000 + 5;
const int MOD = 1e9 + 7;
const int N = 1005;

ll n,k;
int a[2005];

int main()
{
   ios::sync_with_stdio(false); cin.tie(0);cout.tie(0);
   // #ifndef ONLINE_JUDGE 
   //freopen("input.txt", "r", stdin);
   //#endif // ONLINE_JUDGE 
   cin >> n >> k;
   rep(i,1,n) cin >> a[i];
   ll ans = 0;
   rep(i,1,n){
        ll sum = 0;
        rep(j, i + 1, n) {
            if(a[i] > a[j]) sum++;
        }
        ans = (ans + sum * k % MOD * (k + 1) / 2ll % MOD) % MOD;
        sum = 0;
         rep(j, 1, i - 1) {
            if(a[i] > a[j]) sum++;
        }
         ans = (ans + sum * k % MOD * (k - 1) / 2ll % MOD) % MOD;
   }
   cout << ans % MOD << endl;
   return 0;
}