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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
int main(void) {
    long int N,K;
    while(1){
    scanf("%d",&N);
    scanf("%d",&K);
    if(N>0&&K>0){
    	break;
    }
    }
    unsigned int *A,*B;
    int i,k,p;
    long int H=0;
    A=(long int*)malloc(sizeof(long int)*N);
    for(i=0;i<N;i++){
       scanf("%d",&A[i]);
    }
    B=(long int*)malloc(sizeof(long int)*K*N);
    for(k=0;k<K;k++){
    	for(i=k*N;i<(k+1)*N;i++){
       		p=i%N;
       		B[i]=A[p];
    	}
    }
    for(k=0;k<K*N-1;k++){
    	for(i=k+1;i<K*N;i++){
    		if(B[k]>B[i]){
    			H++;
    		}
    	}
    }
    H=H%100000007;
    printf("%d\n",H);
    free(A);
    free(B);
    return 0;
}