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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i=0; i<(int)(n); i++)
#define intt int64_t
#define all(vec) vec.begin(), vec.end()
#define MOD 1000000007

int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    vector<int> vec(n);
    rep(i,n) scanf("%d", &vec.at(i));
    vector<int> small(n, 0);
    vector<int> smalll(n, 0);
    rep(i,n) {
        rep(j,n) {
            if (vec.at(i) > vec.at(j)) {
                small.at(i)++;
                if (i > j) smalll.at(i)++;
            }
        }
    }
    intt ans=0;
    rep(i,n) {
        ans += ((intt)(k+1)/2*small.at(i)-smalll.at(i))*k;
        ans %= MOD;
    }
    printf("%d\n", ans%MOD);
}