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>
#define REP(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)
#define REPS(i, _s, n) for(int i = _s, i##_len = (n); i < i##_len; ++i)
#define ALL(x) (x).begin(), (x).end()
using namespace std;

#define repi(itr, ds) for(auto itr = ds.begin(); itr != ds.end(); itr++)
// vector
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for(T &x : vec) is >> x;
  return is;
}
// pair
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
// vector
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "{";
  for(int i = 0; i < vec.size(); i++) {
    os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
  }
  os << "}\n   ";
  return os;
}
// map
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> &map_var) {
  os << "{";
  repi(itr, map_var) {
    os << *itr;
    itr++;
    if(itr != map_var.end()) os << ", ";
    itr--;
  }
  os << "}";
  return os;
}
// set
template <typename T>
ostream &operator<<(ostream &os, set<T> &set_var) {
  os << "{";
  repi(itr, set_var) {
    os << *itr;
    itr++;
    if(itr != set_var.end()) os << ", ";
    itr--;
  }
  os << "}";
  return os;
}

#define DUMPOUT cerr

void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
  DUMPOUT << head;
  if(sizeof...(Tail) > 0) {
    DUMPOUT << ", ";
  }
  dump_func(std::move(tail)...);
}
#ifdef DEBUG_
#define DEB
#define dump(...)                                                             \
  DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                             \
          << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]" << endl \
          << "    ",                                                          \
      dump_func(__VA_ARGS__)
#else
#define DEB if(false)
#define dump(...)
#endif

typedef long long int ll;
const ll MOD = 1e9 + 7;
const int INF = 1e9;

int main() {
  ll N, K;
  cin >> N >> K;
  auto A = vector<ll>(2001);
  REP(i, N) cin >> A[i];
  // REP(i, 2000) { A[i] = i + 1; }

  auto left = vector<ll>(2001, 0);
  auto right = vector<ll>(2001, 0);
  REP(i, N) {
    REP(j, N) {
      if(A[j] > A[i] && j > i) {
        left[A[j]]++;
      } else if(A[j] > A[i] && i > j) {
        right[A[j]]++;
      }
    }
  }
  ll sum = 0;
  REP(i, N + 1) {
    ll rs = right[i] * (((K + 1) * K / 2) % MOD);
    rs %= MOD;
    ll ls = left[i] * (((K - 1) * K / 2) % MOD);
    ls %= MOD;
    sum += rs;
    sum %= MOD;
    sum += ls;
    sum %= MOD;
  }
  cout << sum << endl;
}