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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using static System.Math;
 
public class Hello{
    public static void Main(){
        long[] read = longa();
        int[] list = inta();
        long fir = 0;
        long sec = 0;
        for(int i=0;i<read[0];i++){
            for(int j=i+1;j<read[0];j++){
                if(list[i] > list[j]){
                    fir++;
                }
            }
        }
        for(int i=0;i<read[0];i++){
            for(int j=0;j<read[0];j++){
                if(list[i] > list[j]){
                    sec++;
                }
            }
        }
        //WriteLine(fir);
        //WriteLine(sec);
        // fir * n + sec * 1/2 * n * n-1
        long ans = 0;
        ans += (fir * read[1]) %1000000007;
        ans += (read[1] * (read[1]-1) /2) %1000000007 * sec;
        WriteLine(ans % 1000000007);

    
    }
    //各種読取
    static string rstr(){
        return ReadLine();
    }
    static int rint(){
        return int.Parse(ReadLine());
    }
    static long rlong(){
        return long.Parse(ReadLine());
    }
    static string[] stra(){
        return ReadLine().Split(' ');
    }
    static int[] inta(){
        string[] read_str_array = ReadLine().Split(' ');
        int[] return_int_array = new int[read_str_array.Length];
        for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_int_array[countup_i] = int.Parse(read_str_array[countup_i]);
        }
        return return_int_array;
    }
    static int[,] inta2(int num_array,int in_array){
        int[,] int_array2 = new int[num_array,in_array];
        for(int i=0;i<num_array;i++){
            int[] temp_array = inta();
            for(int j=0;j<in_array;j++){
                int_array2[i,j] = temp_array[j];
            }
        }
        return int_array2;
    }
    static long[] longa(){
        string[] read_str_array = ReadLine().Split(' ');
        long[] return_long_array = new long[read_str_array.Length];
        for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_long_array[countup_i] = long.Parse(read_str_array[countup_i]);
        }
        return return_long_array;
    }
    static double[] doublea(){
        string[] read_str_array = ReadLine().Split(' ');
        double[] return_double_array = new double[read_str_array.Length];
        for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){
            return_double_array[countup_i] = long.Parse(read_str_array[countup_i]);
        }
        return return_double_array;
    }
    // A/B小数点以下切上
    static long divideup(long divided,long divid){
        long temp_divide = divided/divid;
        if(divided % divid > 0){
            temp_divide++;
        }
        return temp_divide;
    }
    static long GCD(long a,long b){
        if(a < b){
            long temp = a;
            a = b;
            b = temp;
        }
        if(a % b == 0){ return b; }
        else{
            long temp = b;
            b = a%b;
            a = temp;
            return GCD(a,b);
        }
    }
    static long LCM(long a,long b){
        return a * b / GCD(a,b);
    }
}
//知見
// (string) hogestr -> (char[]) hoge
    // char hoge = hogestr.ToCharArray();
// (char[]) hoge -> (string) hogestr
    // string hoge = new string(hoge);
// Hashtable ×  Dictionary ○
// Dictionary hoge<string,int>
    // hoge.ContainsKey(string) -> return boolean
    // hoge[string] -> return int
    // hoge[int] -> CE
// Dictionary hoge<int1,int2>
    // hoge<int1,int2>: hoge[int1] -> return int2
    // KeyValuePair<int,int> k <-> k.Key == int1 / k.Value == int2
// SortedDictionary() > SortedList() ???
// キュー　-> 幅優先 ???
// スタック -> 深さ優先 ???