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: Accepted)

def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y

# mを法とするaの乗法的逆元
def modinv(a, m):
    g, x, y = egcd(a, m)
    if g != 1:
        raise Exception('modular inverse does not exist')
    else:
        return x % m

# nCr mod m
# modinvが必要
# rがn/2に近いと非常に重くなる
def combination(n, r, mod=10 ** 9 + 7):
    r = min(r, n - r)
    res = 1
    for i in range(r):
        res = res * (n - i) * modinv(i + 1, mod) % mod
    return res

n, k = map(int, input().split())
dat = list(map(int, input().split()))
datAfter = [0] * n
datSelf = [0] * n
numAfter = 0
numSelf = 0
m = 10**9 + 7
for i in range(n):
    for j in range(i+1, n):
        if dat[i] > dat[j]:
            numAfter += 1
for i in range(n):
    for j in range(n):
        #print(dat[i], dat[j])
        if dat[i] > dat[j]:
            numSelf += 1
res = 0
res += numAfter * k
res += numSelf * combination(k, 2)
res %= m
#print(numSelf, numAfter, res)
print(res)