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.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;


class Program {

	static long mod = 1000000007;
	//static long mod =998244353;
	static int[] ans;
	static void Main(string[] args) {

		char[] cs = new char[] { ' ' };
		//char[] cs = new char[] { ' ', ',' };

		//int n = int.Parse(Console.ReadLine());
		//int val = int.Parse(Console.ReadLine());
		//int[] vals = parseAry(Console.ReadLine().Split(cs));
		//string str = Console.ReadLine();
		//string[] strs = Console.ReadLine().Split(cs);

		//string res = "";
		long res=0;
		int max=2000;

		int[] vals = parseAry(Console.ReadLine().Split(cs));
		int n = vals[0];
		long kk = vals[1];
		int[] nums = parseAry(Console.ReadLine().Split(cs));


		long plusL = ( ( 1 + kk ) * ( kk ) / 2 ) % mod;
		long plusR = ( ( 1 + kk-1 ) * ( kk-1 ) / 2 ) % mod;


			for ( int i = 0; i < n; i++ ) {

				long cnt = 0;
				long left = 0;
				long right=0;
				for ( int k = 0; k < n; k++ ) {
					if ( k < i ) { if ( nums[i] < nums[k] ) { left++; } }
					if ( k > i ) { if ( nums[i] < nums[k] ) { right++; } }
				}

				long temp = left * plusL % mod;
				long temp2 = right * plusR % mod;

					res += temp+temp2;
				res %= mod;
			}






			Console.WriteLine(res);

			return;

	}



	//	int配列をスペース区切りの文字列に変換
	static string stringBuilderArrayInt(int[] a) {
		StringBuilder sb = new StringBuilder();

		for ( int i = 0; i < a.Length; i++ ) {
			if ( a[i] == 1 ) { sb.Append(i + " "); }
		}
		if ( sb.Length > 1 ) {
			sb = sb.Remove(sb.Length - 1, 1);
		}
		return sb.ToString(); ;
	}


	//	文字列strをindex番目からdで上書き
	static string strOverWrite(string str, int i, string s) {
		return str.Remove(i, s.Length).Insert(i, s);
	}

	//　String[]  ----->  int[]
	static int[] parseAry(string[] str) {
		int[] nums = new int[str.Length];
		for ( int i = 0; i < str.Length; i++ ) { nums[i] = int.Parse(str[i]); }
		return nums;
	}
	//　String[]  ----->  long[]
	static long[] parseAryL(string[] str) {
		long[] nums = new long[str.Length];
		for ( int i = 0; i < str.Length; i++ ) { nums[i] = long.Parse(str[i]); }
		return nums;
	}

	//文字列を1文字ずつString[]に入れる
	static string[] strAry(String str) {
		string[] re = new string[str.Length];
		for ( int i = 0; i < str.Length; i++ ) { re[i] = str[i].ToString(); }
		return re;
	}


	
}