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 Python (3.4.3) (judge verdict: Wrong Answer)

from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush
from fractions import gcd
from copy import deepcopy
import math
import queue
#import numpy as np
Mod = 1000000007




def main(): #startline-------------------------------------------
    n,k = map(int,input().split())
    a = list(map(int,input().split()))
    ans = 0
    for i in range(n):
        for j in range(n-1,i,-1):
            if a[j] < a[j-1]:
                a[j],a[j-1] = a[j-1],a[j]
                ans += 1
    ans %= Mod
    u = Counter(a)
    ul = list(u.keys())
    ul.sort(reverse = True)
    p = len(ul)   
    ans *= k
    ans %= Mod
    temp = 0
    for i in range(p):
        temp += u[ul[i]]
        ans += (n-temp)*(k*(k-1)//2)
        ans %= Mod
    print(ans)

if __name__ == "__main__":
    main() #endline===============================================





def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n < 2 is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))):
        if prime[i] == 1:
            for j in range(2*i,n):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n):
        if prime[i] == 1:
            res.append(i)
    return res

class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[py] = px
        else:
            self.parent[px] = py

    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)