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.
n, k = map(int, input().split())
a = list(map(int, input().split()))
 
mod = 10 ** 9 + 7

b = [[0, 0] for _ in range(n)]
for i in range(n):
    for j in range(n):
        if a[i] > a[j]:
            if i > j:
                b[i][0] += 1
            else:
                b[i][1] += 1
ans = 0
sigmakp = int(((k * (k+1)) % mod) / 2) % mod
sigmakn = int(((k * (k-1)) % mod) / 2) % mod
for i in range(n):
    mae = (sigmakn * b[i][1]) % mod
    ushiro = (sigmakp * b[i][0]) % mod
    ans = ((mae + ushiro) % mod + ans) % mod 
print(ans)
n, k = map(int, input().split())
a = list(map(int, input().split()))
 
mod = 10 ** 9 + 7

b = [[0, 0] for _ in range(n)]
for i in range(n):
    for j  in range(i,n):
        if a[i] > a[j]:
                b[i][1] += 1
        elif a[i] < a[j]:
        		b[i][0] +=1


b[i][0]=b[i][0]%mod		
b[i][1]=b[i][1]%mod	
ans = 0
sigmakp = ((k * (k+1) ) // 2) % mod

sigmakn = ((k * (k-1)) // 2) % mod

for i in range(n):
    mae = int(sigmakn * b[i][0]) % mod
    ushiro = int(sigmakp * b[i][1]) % mod
    ans = int((mae + ushiro) + ans) % mod 
print(ans)
