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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

#define a 1000000000+7

void swap (int *x, int *y) {
  int temp;    // 値を一時保存する変数

  temp = *x;
  *x = *y;
  *y = temp;
}

/***
* pivotを決め、
* 全データをpivotを境目に振り分け、
* pivotの添え字を返す
***/
int partition (int array[], int left, int right) {
  int i, j, pivot;
  i = left;
  j = right + 1;
  pivot = left;   // 先頭要素をpivotとする

  do {
    do { i++; } while (array[i] < array[pivot]);
    do { j--; } while (array[pivot] < array[j]);
    // pivotより小さいものを左へ、大きいものを右へ
    if (i < j) { swap(&array[i], &array[j]); }
  } while (i < j);

  swap(&array[pivot], &array[j]);   //pivotを更新

  return j;
}

/* クイックソート */
void quick_sort (int array[], int left, int right) {
  int pivot;

  if (left < right) {
    pivot = partition(array, left, right);
    quick_sort(array, left, pivot-1);   // pivotを境に再帰的にクイックソート
    quick_sort(array, pivot+1, right);
  }
}

int main(){
    long long N,K;
    long long counta=0,ct=0;
    long long  ans=0;
    scanf("%lld%lld",&N,&K);
    long long A[N],count[N];
    for(int i=0;i<N;i++){
        count[i]=1; 
    }
    for(int i=0;i<N;i++){
        scanf("%lld",&A[i]);
    }
    for(int i=0;i<N;i++){
        for(int j=0;j<i;j++){
            if(A[i]<A[j]){
                ct++;
            }
        }
    }
    quick_sort(A,0,N-1);
    for(int i=0;i<N;i++){
            if(A[i]==A[i+1]){
                count[counta]++;
            }else{
                counta++;
            }
    }
    long long eq=0;
    for(int i=0;i<counta;i++){
        eq=eq+(count[i]*count[i])%a;
    }
    ans=((N*N)%a-eq+ct)%a;
    ans*=((K/2)*(K-1))%a;
    printf("%lld",ans%a);
        return 0;
}