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>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
using namespace std;
#define oset tree<long long, null_type,less_equal<long long>, rb_tree_tag,tree_order_statistics_node_update>
using ll = long long;
ll mod=1e9+7;




int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    t = 1;
    while(t--){
        ll n,k,i;
        cin>>n>>k;
        oset s,s2;
        ll a[n+2];
        for(i=0;i<n;i++){
            cin>>a[i];
            s2.insert(a[i]);
        }
        
        ll bigonR[n+2],bigger[n+2];
        for(i=0;i<n;i++){
            bigger[i] = s2.order_of_key(a[i]);
        }
        
        
        for(i=n-1;i>=0;i--){
            bigonR[i]=s.order_of_key(a[i]);
            s.insert(a[i]);
            
        }
        ll sum=0;
        for(i=0;i<n;i++){
            // cout << bigger[i] << ' ' << bigonR[i] << "\n";
            ll kk = (k * (k - 1)) / 2;
            kk %= mod;
            sum += ((bigger[i] * kk) % mod);
            sum %= mod;
            sum += ((bigonR[i] * k) % mod);
            sum %= mod;
        }
        cout << sum;
        cout<<"\n";
    }
}