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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")


void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a=c;}
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll minn(ll n, ll a[n]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[n]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll ABS(ll a){return a>=0?a:(-a);}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

typedef struct{
    ll aa;
    ll bb;
}frequent;
// (mod MOD1)
ll mod(ll n){return n%=MOD1;}
ll inv[200005],finv[200005];
void makeinv(ll n){
    inv[1]=1;finv[0]=finv[1]=1;
    rep(i,2,n+1){
        inv[i]=MOD1 - inv[MOD1%i] * (MOD1 / i) % MOD1;
        finv[i] = finv[i - 1] * inv[i] % MOD1;
    }
    // for(ll i=2;i<=n;i++){
    //     inv[i]=MOD1 - inv[MOD1%i] * (MOD1 / i) % MOD1;
    //     finv[i] = finv[i - 1] * inv[i] % MOD1;
    // }
}

int main(void){
    // char ;
    makeinv(2);
    ll n,k,ans=0;
    scanf("%lld%lld",&n,&k);
    ll a[n],cnt[2000]={},inv2[2000]={};
    rep(i,0,n){
        scanf("%lld",&a[i]);
        cnt[a[i]-1]++;
    }
    ll cum[2000]={};
    cum[1998]=cnt[1999];
    rrep(i,1997,0){
        cum[i]=cum[i+1]+cnt[i+1];
    }
    rep(i,0,n-1){
        rep(j,i+1,n){
            if(a[i]>a[j]){
                inv2[a[j]-1]++;
            }
        }
    }
    rep(i,0,maxn(n,a)){
        ans+= inv2[i] * k +  mod( mod(cum[i] *inv[2]) * mod( k* (k-1) ) ) ;
        ans%=MOD1;
    }
    // printf("%lld ",inv[2]);

    printf("%lld\n",ans);
    return 0;
}