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: Runtime Error)

# -*- coding: utf-8 -*-
# 
import math
import sys
import itertools
import numpy as np
 
DIV_NUM = 1000000000 + 7
 
INPUT_NUMS = list(map(int, input().split()))
N=int(INPUT_NUMS[0])
K=int(INPUT_NUMS[1])
 
A = [int(x) for x in input().split()]
AA = np.array(A)
if N==1:
    print(0)
    exit()
 
tree = np.array([0] * (N+1) )
 
# https://atcoder.jp/contests/chokudai_S001/submissions/6789704
def BIT_update(tree,x):
    while x <= N:
        tree[x] += 1
        x += x&(-x)
def BIT_sum(tree,x):
    s = 0
    while x:
        s += tree[x]
        x -= x&(-x)
    return s
 
answer = N*(N-1)//2
for x in AA:
    answer -= BIT_sum(tree,int(x))
    BIT_update(tree,int(x))
 
B = sorted(A)
 
q=0
p=B[0]
r=0
for i in range(1,N):
    if B[i]>p:
        p=B[i]
        r=i
    q += r
 
# print(answer,K)
answer = answer % DIV_NUM
answer = int(answer)
answer = (answer * K) % DIV_NUM
q = q % DIV_NUM
answer = answer + q * K * (K-1) // 2
answer = answer % DIV_NUM
print(answer)