#include <bits/stdc++.h>

using namespace std;

#define int long long
// #define double long double
#define FOR(i, a, b) for(ll i = (a); i < (b); ++i)
#define FORR(i, a, b) for(ll i = (a); i > (b); --i)
#define REP(i, n) for(ll i = 0; i < (n); ++i)
#define REPR(i, n) for(ll i = n; i >= 0; i--)
#define FOREACH(x, a) for(auto &(x) : (a))
#define VECCIN(x)                                                              \
    for(auto &youso_ : (x)) cin >> youso_
#define bitcnt __builtin_popcount
#define SZ(x) ((ll)(x).size())
#define fi first
#define se second
#define All(a) (a).begin(), (a).end()
#define rAll(a) (a).rbegin(), (a).rend()
template <typename T = long long> inline T IN() {
    T x;
    cin >> x;
    return (x);
}
inline void CIN() {}
template <class Head, class... Tail>
inline void CIN(Head &&head, Tail &&... tail) {
    cin >> head;
    CIN(move(tail)...);
}
#define CCIN(...)                                                              \
    char __VA_ARGS__;                                                          \
    CIN(__VA_ARGS__)
#define DCIN(...)                                                              \
    double __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define LCIN(...)                                                              \
    ll __VA_ARGS__;                                                            \
    CIN(__VA_ARGS__)
#define SCIN(...)                                                              \
    string __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define Printv(v)                                                              \
    {                                                                          \
        REP(x, v.size()) { cout << v[x] << (x == v.size() - 1 ? "\n" : " "); } \
    }
template <typename T = string> inline void eputs(T s) {
    cout << s << "\n";
    exit(0);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *)array, (T *)(array + N), val);
}
template <typename T> using PQG = priority_queue<T, vector<T>, greater<T>>;
template <typename T> using PQ = priority_queue<T>;

typedef long long ll;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<ll, ll> PL;
typedef vector<PL> VPL;

const int INF = 1e9;
const int MOD = 1e9 + 7;
const double PI = atan(1.0) * 4.0;
const ll LINF = 1e18;
const ll dx[] = {1, 0, -1, 0};
const ll dy[] = {0, 1, 0, -1};

template <int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if(val < 0) v += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator-() const noexcept { return val ? MOD - val : 0; }
    constexpr Fp operator+(const Fp &r) const noexcept {
        return Fp(*this) += r;
    }
    constexpr Fp operator-(const Fp &r) const noexcept {
        return Fp(*this) -= r;
    }
    constexpr Fp operator*(const Fp &r) const noexcept {
        return Fp(*this) *= r;
    }
    constexpr Fp operator/(const Fp &r) const noexcept {
        return Fp(*this) /= r;
    }
    constexpr Fp &operator+=(const Fp &r) noexcept {
        val += r.val;
        if(val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp &operator-=(const Fp &r) noexcept {
        val -= r.val;
        if(val < 0) val += MOD;
        return *this;
    }
    constexpr Fp &operator*=(const Fp &r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp &operator/=(const Fp &r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while(b) {
            long long t = a / b;
            a -= t * b;
            swap(a, b);
            u -= t * v;
            swap(u, v);
        }
        val = val * u % MOD;
        if(val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator==(const Fp &r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator!=(const Fp &r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream &operator<<(ostream &os,
                                         const Fp<MOD> &x) noexcept {
        return os << x.val;
    }
    friend constexpr istream &operator>>(istream &is, Fp<MOD> &x) noexcept {
        return is >> x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if(n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if(n & 1) t = t * a;
        return t;
    }
};

// new
template <typename T, typename E> struct SegmentTree {
    using F = function<T(T, T)>;
    using G = function<T(T, E)>;
    int n;
    F f;
    G g;
    T d1;
    E d0;
    vector<T> dat;
    SegmentTree(){};
    SegmentTree(int n_, F f, G g, T d1, vector<T> v = vector<T>())
        : f(f), g(g), d1(d1) {
        init(n_);
        if(n_ == (int)v.size()) build(n_, v);
    }
    void init(int n_) {
        n = 1;
        while(n < n_) n *= 2;
        dat.clear();
        dat.resize(2 * n - 1, d1);
    }
    void build(int n_, vector<T> v) {
        for(int i = 0; i < n_; i++) dat[i + n - 1] = v[i];
        for(int i = n - 2; i >= 0; i--)
            dat[i] = f(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
    void update(int k, E a) {
        k += n - 1;
        dat[k] = g(dat[k], a);
        while(k > 0) {
            k = (k - 1) / 2;
            dat[k] = f(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }
    inline T query(int a, int b) {
        T vl = d1, vr = d1;
        for(int l = a + n, r = b + n; l < r; l >>= 1, r >>= 1) {
            if(l & 1) vl = f(vl, dat[(l++) - 1]);
            if(r & 1) vr = f(dat[(--r) - 1], vr);
        }
        return f(vl, vr);
    }
};

using mint = Fp<MOD>;
ll val[2000];

signed main() {
    LCIN(N, K);
    VL A(N);
    VECCIN(A);
    VL tmp = A;
    sort(All(tmp));
    REP(i, N) { val[i] = tmp.end() - upper_bound(All(tmp), A[i]); }
    SegmentTree<ll, ll> num(2001, [](ll a, ll b) { return a + b; },
                            [](ll a, ll b) { return a + b; }, 0, VL(2001, 0));
    mint ans = 0, sum = 0;
    REP(i, N) {
        sum += i - num.query(0, A[i] + 1);
        num.update(A[i], 1);
    }
    ans += sum * K;
    REP(i, N) ans += (mint)K * (K - 1) / 2 * val[i];
    cout << ans << "\n";
}


Measured memory_bytes: 384