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: Runtime Error)

#include <bits/stdc++.h>
#define REP(i,n) for(int64_t i=0;i<(n);i++)
#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REV(c) reverse((c).begin(),(c).end())
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define Yes cout << "Yes" << endl;
#define No cout << "No" << endl;
#define VIN(n,v) for(int64_t i=0;i<(n);i++){cin >> (v).at(i);}

using namespace std;
typedef int64_t I;
typedef pair<I,I> P;
typedef vector<I> V;

I MOD = pow(10,9)+7;

// 最大公約数
I gcd(I a, I b) {
  return b != 0 ? gcd(b, a % b) : a;
}

// 最小公倍数
I lcm(I a, I b) {
  return a * b / gcd(a, b);
}

int ctoi(char c){
  return c - '0';
}

int main(){
  I n, k;
  cin >> n >> k;
  V a(n);
  VIN(n,a);
  if (n == 1){
    cout << 0 << endl;
    return 0;
  }
  vector<P> an(n);
  REP(i,n){
    an.at(i).first = 0;
    an.at(i).second = 0;
  }
  REP(i,k){
    REP(j,k){
      an.at(i).first = k;
    }
  }
  REP(i,n){
    REP(j,n){
      if (i < j && a.at(i) > a.at(j)){
        an.at(i).second++;
      }
      if (i > j && a.at(i) > a.at(j)){
        an.at(i).first++;
      }
    }
  }
  I t=0;
  REP(i,n){
    t += ((k + 1) * k / 2) * an.at(i).second + ((k) * (k-1) / 2 ) * an.at(i).first;
    t %= MOD;
  }
  cout << t << endl;
  return 0;
}