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)

/*
これを入れて実行
g++ code.cpp
./a.out
 */

#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <cmath>
#include <math.h>
#include <tuple>
#include <iomanip>

using namespace std;
typedef long long ll;
typedef long double ld;

int dy4[4] = {-1, 0, +1, 0};
int dx4[4] = {0, +1, 0, -1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dx8[8] = {0, 1, 1, 1, 0, -1, -1, -1};

const long long INF = 1LL << 60;
const ll MOD = 1e9 + 7;

bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s){
    return f.second > s.second;
}

ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a % b);
}

ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll nCr(ll n, ll r){
    if(r == 0 || r == n){
        return 1;
    } else if(r == 1){
        return n;
    }
    return (nCr(n - 1, r) + nCr(n - 1, r - 1));
}

ll nPr(ll n, ll r){
    r = n - r;
    ll ret = 1;
    for (ll i = n; i >= r + 1; i--) ret *= i;
    return ret;
}

//-----------------------ここから-----------

int main(void){
    ll n, k;
    cin >> n >> k;
    vector<ll> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];

    vector<ll> l(n, 0);

    for(int i = 0; i < n; i++){
        for(int j = 0; j < i; j++){
            if(a[i] < a[j]){
                l[i]++;
            }
        }
    }

    ll bucket[2100] = {0};

    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(a[i] < a[j]){
                bucket[a[i]]++;
            }
        }
    }

    // ll ans = 0;
    // vector<ll> sum(n, 0);
    // for(int i = 0; i < k; i++){
    //     for(int j = 0; j < n; j++){
    //         if(i == 0){
    //             sum[j] += l[j];
    //             ans += l[j];
    //             continue;
    //         }

    //         ans += bucket[a[j]] + sum[j];
    //         sum[j] = bucket[a[j]] + sum[j];
    //     }
    // }
    // cout << ans % MOD << endl;
    
    ll tmp = 0;
    ll ftmp = 0;

    for(int i = 0; i < n; i++){
        ftmp += l[i];
    }
    for(int i = 0; i < n; i++){
        tmp += bucket[a[i]];
    }
    
    ll ans = 0;
    ll last = 0;
    for(int i = 0; i < k; i++){
        if(i == 0){
            ans += ftmp;
            last = ftmp;
            continue;
        }
        ans += last + tmp;
        ll t2 = last + tmp;
        last = t2;
        last %= MOD;
        ans %= MOD;
    }
    cout << ans % MOD << endl;
}