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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.math.BigInteger;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        long k = in.nextLong();
        int[] a = new int[n];
        for(int i=0; i<n; i++) {
            a[i] = in.nextInt();
        }
        SortedMap<Integer, Integer> m = new TreeMap<>();
        long invPerA = 0;
        for(int i=0; i<n; i++) {
            invPerA += countGreater(m, a[i]);
            int c = m.containsKey(a[i]) ? m.get(a[i]) : 0;
            m.put(a[i], c+1);
        }
        long invForPrevA = 0;
        for(int i=0; i<n; i++) {
            invForPrevA += countGreater(m, a[i]);
        }
        BigInteger x1 = new BigInteger(Long.toString(invPerA));
        BigInteger x2 = new BigInteger(Long.toString(invForPrevA));
        BigInteger k2 = new BigInteger(Long.toString(k));
        BigInteger k3 = new BigInteger(Long.toString(k-1));
        BigInteger two = new BigInteger("2");
        BigInteger c = x1.multiply(k2).add(x2.multiply(k2).multiply(k3).divide(two));
        //long c = invPerA * k + invForPrevA * (k * (k-1) / 2);
        BigInteger mod = new BigInteger("1000000007");
        c = c.mod(mod);
        System.out.println(c.toString());
    }
    private static int countGreater(SortedMap<Integer, Integer> m, int key) {
        SortedMap<Integer, Integer> sub = m.tailMap(key);
        Set<Integer> keySet = sub.keySet();
        Iterator<Integer> it = keySet.iterator();
        int c = 0;
        while(it.hasNext()) {
            int k = it.next();
            if(k == key) {
                continue;
            }
            c += sub.get(k);
        }
        return c;
    }
}