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 ll long long
using namespace std;
int n;
const ll Mod=1000000007ll;
ll k,kk,ans=0;
ll a[2010];
ll c1[2010],c2[2010];
int main() {
	cin>>n>>k;
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++) //总 
		for(int j=1;j<=n;j++) {
			if(i==j) continue;
			if(a[i]>a[j]) c1[i]++;
			c1[i]%Mod;
		}
	for(int i=1;i<=n;i++) //后 
		for(int j=i+1;j<=n;j++) {
			if(a[i]>a[j]) c2[i]++;
			c2[i]%Mod;
		}
//	for(int i=1;i<=n;i++)
//		printf("%d ",c1[i]);
//	printf("\n");
//	for(int i=1;i<=n;i++)
//		printf("%d ",c2[i]);
//	printf("\n");
	k%=Mod;
	kk=(1+k)*k/2;kk--;
	kk%=Mod;
	for(int i=1;i<=n;i++) {
		ans+=c1[i]*kk%Mod;
		ans%=Mod; 
	}
	for(int i=1;i<=n;i++) {
		ans+=c2[i];
		ans%=Mod; 
	}
	printf("%lld",ans);
	return 0;
}