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: Wrong Answer)

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

public class Main {

    static String br = System.getProperty("line.separator");
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int N = nextInt();
        long K = nextLong();

        int[] A = new int[N];
        for(int i=0; i<N; i++){
            A[i] = nextInt();
        }

        long count = 0;
        for(int i=0; i<N-1; i++){
            int before = A[i];
            for(int j=i+1; j<N; j++) {
                if(before > A[j]) count++;
            }
        }

        long sum = 0;
        for(long k=K; k>=1; k--){
            sum += count*k%1000000007L;
        }

        long result = sum%(1000000007L);
        println(result);
    }

    //最小公倍数lcm
    static long lcm (long a, long b) {
        long temp;
        long c = a;
        c *= b;
        while((temp = a%b)!=0) {
            a = b;
            b = temp;
        }
        return (c/b);
    }

    static int nextInt() {
        return Integer.parseInt(next());
    }

    static Long nextLong() {
        return Long.parseLong(next());
    }

    static String next() {
        return sc.next();
    }

    static String nextLine(){
        return sc.nextLine();
    }

    static void println(Object o) {
        out.println(o);
        out.flush();
    }

    static void print(Object o) {
        PrintWriter out = new PrintWriter(System.out);
        out.print(o);
        out.flush();
    }
}