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.*;
import java.util.*;

/**
 * @author masayamatu
 * 
 */
public class Main {
  
  public static void main(String[] args) throws IOException{
    FastScanner sc = new FastScanner();
    int n = sc.nextInt();
    long k = sc.nextLong();
    long[] A = new long[n];
    long gyaku = 0;
    long ans1 = 0;
    long ans2 = 0;
    long jun = 0;
    for(int i = 0; i < n; i++) {
      A[i] = sc.nextInt();
    }
    for(int i = 0; i < n; i++) {
      for(int j = i+1; j < n; j++) {
        if(A[i] > A[j]) {
          gyaku++;
        }else if(A[i] < A[j]) {
          jun++;
        }
      }
    }
    ans1 = (gyaku*k)%1000000007;
    ans2 = ((gyaku+jun)%1000000007)*((k*((k-1)/2))%1000000007);
    System.out.println((ans1+ans2)%1000000007);
 }
 
}
class FastScanner {
  private final InputStream in = System.in;
  private final byte[] buffer = new byte[1024];
  private int ptr = 0;
  private int buflen = 0;
  private boolean hasNextByte() {
      if (ptr < buflen) {
          return true;
      }else{
          ptr = 0;
          try {
              buflen = in.read(buffer);
          } catch (IOException e) {
              e.printStackTrace();
          }
          if (buflen <= 0) {
              return false;
          }
      }
      return true;
  }
  private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
  private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
  public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
  public String next() {
      if (!hasNext()) throw new NoSuchElementException();
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while(isPrintableChar(b)) {
          sb.appendCodePoint(b);
          b = readByte();
      }
      return sb.toString();
  }
  public long nextLong() {
      if (!hasNext()) throw new NoSuchElementException();
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
          minus = true;
          b = readByte();
      }
      if (b < '0' || '9' < b) {
          throw new NumberFormatException();
      }
      while(true){
          if ('0' <= b && b <= '9') {
              n *= 10;
              n += b - '0';
          }else if(b == -1 || !isPrintableChar(b)){
              return minus ? -n : n;
          }else{
              throw new NumberFormatException();
          }
          b = readByte();
      }
  }
  public int nextInt() {
      long nl = nextLong();
      if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
      return (int) nl;
  }
  public double nextDouble() { return Double.parseDouble(next());}
}