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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vl = vector<ll>;
using vll = vector<vl>;
using Pll = pair<ll, ll>;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define all(v) v.begin(), v.end()
#define sz(x) ((int) x.size())
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define F first
#define S second
const int MOD = 1e9+7;
const ll INF = 2e15;
template<class T> void print(const T& t){ cout << t << endl; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

int main(){
  ll n, k;
  cin >> n >> k;
  vl a(n);
  rep(i,n){
    cin >> a[i];
  }
  ll sum = 0;
  rep(i,n){
    ll p = 0;
    ll q = 0;
    for (ll j = i + 1; j < n; j++){
      if(a[i]>a[j]){
        p++;
      }
    }
    for (ll j = 0; j < i;j++){
      if(a[i]>a[j]){
        q++;
      }
    }
    cout << p << " " << q << endl;
    sum += p * k * (k + 1) / 2;
    sum += q * k * (k - 1) / 2;
  }
  sum %= MOD;
  print(sum);
}