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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define INF_LL (int64)1e18
#define INF (int32)1e9
#define REP(i, n) for(int64 i = 0;i < (n);i++)
#define FOR(i, a, b) for(int64 i = (a);i < (b);i++)
#define all(x) x.begin(),x.end()
#define fs first
#define sc second

using int32 = int_fast32_t;
using uint32 = uint_fast32_t;
using int64 = int_fast64_t;
using uint64 = uint_fast64_t;
using PII = pair<int32, int32>;
using PLL = pair<int64, int64>;

const double eps = 1e-10;

template<typename A, typename B>inline void chmin(A &a, B b){if(a > b) a = b;}
template<typename A, typename B>inline void chmax(A &a, B b){if(a < b) a = b;}

template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}

template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}

template<typename T,typename U,typename... V>
typename enable_if<is_same<T, U>::value!=0>::type
fill_v(U &u,const V... v){u=U(v...);}

template<typename T,typename U,typename... V>
typename enable_if<is_same<T, U>::value==0>::type
fill_v(U &u,const V... v){
  for(auto &e:u) fill_v<T>(e,v...);
}

template<::std::uint_fast64_t mod>
class ModInt{
private:
    using value_type = ::std::uint_fast64_t;
    value_type n;
public:
    ModInt() : n(0) {}
    ModInt(value_type n_) : n(n_ % mod) {}
    ModInt(const ModInt& m) : n(m.n) {}

    template<typename T>
    explicit operator T() const { return static_cast<T>(n); }
    value_type get() const { return n; }

    friend ::std::ostream& operator<<(::std::ostream &os, const ModInt<mod> &a) {
        return os << a.n;
    }

    friend ::std::istream& operator>>(::std::istream &is, ModInt<mod> &a) {
        value_type x;
        is >> x;
        a = ModInt<mod>(x);
        return is;
    }

    bool operator==(const ModInt& m) const { return n == m.n; }
    bool operator!=(const ModInt& m) const { return n != m.n; }
    ModInt& operator*=(const ModInt& m){ n = n * m.n % mod; return *this; }

    ModInt pow(value_type b) const{
        ModInt ans = 1, m = ModInt(*this);
        while(b){
            if(b & 1) ans *= m;
            m *= m;
            b >>= 1;
        }
        return ans;
    }

    ModInt inv() const { return (*this).pow(mod-2); }
    ModInt& operator+=(const ModInt& m){ n += m.n; n = (n < mod ? n : n - mod); return *this; }
    ModInt& operator-=(const ModInt& m){ n += mod - m.n; n = (n < mod ? n : n - mod); return *this; }
    ModInt& operator/=(const ModInt& m){ *this *= m.inv(); return *this; }
    ModInt operator+(const ModInt& m) const { return ModInt(*this) += m; }
    ModInt operator-(const ModInt& m) const { return ModInt(*this) -= m; }
    ModInt operator*(const ModInt& m) const { return ModInt(*this) *= m; }
    ModInt operator/(const ModInt& m) const { return ModInt(*this) /= m; }
    ModInt& operator++(){ n += 1; return *this; }
    ModInt& operator--(){ n -= 1; return *this; }
    ModInt operator++(int){
        ModInt old(n);
        n += 1;
        return old;
    }
    ModInt operator--(int){
        ModInt old(n);
        n -= 1;
        return old;
    }
    ModInt operator-() const { return ModInt(mod-n); }
};

constexpr int64 mod = 1e9+7;
using Mint = ModInt<mod>;
int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int64 N, K;
	cin >> N >> K;
	vector<int64> A(N);
	REP(i, N) cin >> A[i];
	vector<Mint> cnt(2001, 0);
	Mint one = 0;
    REP(i, N) {
        cnt[A[i]]++;
        FOR(j, A[i]+1, 2001) {
            one += cnt[j];
        }
    }
    Mint res = one * K;
    Mint sum = 0;
    for (int64 i = 2000; i >= 0; i--) {
        res += cnt[i] * sum * Mint(K-1) * Mint(K) / Mint(2);
        sum += cnt[i];
    }
    cout << res << endl;
}