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 <iostream>
#include <vector>
#include <set>
#include <queue>
#include <map>
#include <cassert>
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define RREP(i,n) for(int i=(n)-1; i>=0; --i)
#define SREP(i,s,n) for(int i=(s), i##_len=(n); i<i##_len; ++i)


using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

template <typename T>
T gcd(T a, T b) {
    if(a < b) gcd(b, a);
    if (b == 1) return 1;
    T r;
    while ((r=a%b)) {
        a = b;
        b = r;
    }
    return b;
}

bool comp(pair<int, int> a, pair<int, int> b) {
    return a.second > b.second;
}

struct UnionFind {
    vector<int> par;
    vector<int> rank;
    
    UnionFind(int n = 1) {
        init(n);
    }
    
    void init(int n = 1) {
        par.resize(n); rank.resize(n);
        for (int i = 0; i < n; ++i) {
            par[i] = i;
            rank[i] = 0;
        }
    }
    
    int root(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            return par[x] = r;
        }
    }
    
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
    
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        return true;
    }
};

#define MOD (1000000007)

int main(int argc, const char * argv[]) {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    ull N, K;
    cin >> N >> K;
    
    vector<int> A(N);
    
    ull cnt = 0;
    REP(i, N) {
        cin >> A[i];
    }
    REP(i, N) {
        ull a = 0, b = 0;
        REP(j, N) {
            if (A[i]>A[j]) a++;
            if (A[i]>A[j] && i<j) b++;
        }
        cnt = ((cnt + (K*b)%MOD)%MOD + ((((K-1)*K)%MOD)*a/2)%MOD)%MOD;
    }
    
    cout << cnt << endl;
}