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.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		const long MOD = 1000000007;
		static void Main(string[] args)
		{
			Scanner cin = new Scanner();
			long n = cin.Long();
			long k = cin.Long();
			long[] a = cin.ArrayLong((int)n);
			long[] pre = new long[n];
			long[] post = new long[n];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < n; j++) {
					if (i == j) {
						continue;
					}

					if (j < i) {
						if (a[i] > a[j]) {
							pre[i]++;
						}
					} else {
						if (a[i] > a[j]) {
							post[i]++;
						}
					}
				}
			}

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

				ans = (ans + (((k * (k + 1) / 2) % MOD) * post[i] % MOD)) % MOD;
				ans = (ans + (((k * (k - 1) / 2) % MOD) * pre[i] % MOD)) % MOD;
			}

			Console.WriteLine(ans);
		}
	}

	public class Scanner
	{
		private readonly char[] delimiter_ = new char[] { ' ' };
		private string[] buf_;
		private int index_;

		public Scanner()
		{
			buf_ = new string[0];
			index_ = 0;
		}

		public string Next()
		{
			if (index_ < buf_.Length) {
				return buf_[index_++];
			}

			string st = Console.ReadLine();
			while (st == "") {
				st = Console.ReadLine();
			}

			buf_ = st.Split(delimiter_, StringSplitOptions.RemoveEmptyEntries);
			if (buf_.Length == 0) {
				return Next();
			}

			index_ = 0;
			return buf_[index_++];
		}

		public int Int()
		{
			return int.Parse(Next());
		}
		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Int() + add;
			}
			return Array;
		}

		public long Long()
		{
			return long.Parse(Next());
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Long() + add;
			}
			return Array;
		}

		public double Double()
		{
			return double.Parse(Next());
		}

		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Double() + add;
			}
			return Array;
		}
	}
}