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 <cmath>
#include <cstdio>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_set>
#include <vector>

using namespace std;

#define FOR(i, N) for (int i = 0; i < (int)N; i++)
#define FORIN(i, a, b) for (int i = a; i < (int)b; i++)
#define ALL(x) (x).begin(), (x).end()
#define MOD 1000000007
#define DEBUG(...) debug(__LINE__, ":" __VA_ARGS__)

using Pi = pair<int, int>;
using ll = long long;
const int INF = 1 << 28;

string to_string(string s) {
  return s;
}

template<class S,class T>
string to_string(pair<S,T> p){
	return "{"+to_string(p.first)+","+to_string(p.second)+"}";
}

template <class T>
string to_string(vector<T> v) {
  string ret = "{";
  for (int i = 0; i < v.size() - 1; ++i) {
    ret += to_string(v[i]) + ",";
  }
  if (v.size() > 0) {
    ret += to_string(v.back());
  }
  ret += "}";
  return ret;
}

void debug() {
  cerr << endl;
}

template <class Head, class... Tail>
void debug(Head head, Tail... tail) {
  cerr << to_string(head) << " ";
  debug(tail...);
}

void print() {
  cout << endl;
}

template <class Head, class... Tail>
void print(Head head, Tail... tail) {
  cout << to_string(head);
  print(tail...);
}

void get() {}

template <class Head, class... Tail>
void get(Head& head, Tail&... tail) {
  cin >> head;
  get(tail...);
}

template <class T>
void getv(vector<T>& vec) {
  for (int i = 0; i < vec.size(); ++i)
    cin >> vec[i];
}

int main() {
	ll N,K;
	cin>>N>>K;
	vector<int> A(N);
	FOR(i,N)cin>>A[i];
	vector<int> dec_after(N),dec_all(N);
	for(int i=0;i<N;++i){
		for(int j=i+1;j<N;++j){
			if(A[i]>A[j])dec_after[i]++;
		}
	}
	for(int i=0;i<N;++i){
		for(int j=0;j<N;++j){
			if(A[i]>A[j])dec_all[i]++;
		}
	}
	ll ans=0;
	for(int i=0;i<N;++i){
		ans=(ans+dec_after[i]*K)%MOD;
		ans=(ans+dec_all[i]*K*(K-1)/2)%MOD;
	}
	print(ans);
  return 0;
}