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>
typedef struct{
    long long front;
    long long rear;
}set;

int main()
{   int n,k;
    scanf("%d%d",&n,&k);
    int a[n];
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    set b[n];
    for(int i=0;i<n;i++){
        b[i].front=0;
        b[i].rear=0;
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<i;j++){
            if(a[i]<a[j]){
                b[i].front++;
                //printf("b[%d].f increased @%d\n",i,j);
            }
        }for(int j=i+1;j<n;j++){
            if(a[i]<a[j]){
                b[i].rear++;
                //printf("b[%d].r increased @%d\n",i,j);
            }
        }
    }
    long long ans=0;
    for(int i=0;i<n;i++){
        long long tmp=b[i].front*(k+1);
        tmp%=1000000007;
        tmp*=k/2;
        tmp%=1000000007;
        ans+=tmp;
        ans%=1000000007;
        tmp=b[i].rear*(k-1);
        tmp%1000000007;
        tmp*=k/2;
        tmp%1000000007;
        ans+=tmp;
        ans%=1000000007;
    }
    

    
    printf("%lld",ans);

    return 0;
}