Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

* Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

* Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.




Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

* 2≤N≤2,000
* a_i are integers.
* 1≤a_i≤10^9
* x is an integer.
* 1≤x≤10^9

Input

The input is given from Standard Input in the following format:


N x
a_1 a_2 ... a_N


Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Examples

Input

2 10
1 100


Output

12


Input

3 10
100 1 100


Output

23


Input

4 10
1 2 3 4


Output

10
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
 
bool SortComp(pair<long long int, int> p1, pair<long long int, int> p2)
{
	if(p1.first < p2.first)
		return true;
	else
		return false;
}
 
int main(void)
{
	int num;
	long long int changet, catcht[2010]= {}, maxchange= 0, sum= 0;
	pair<long long int, int> timecheck[2010];
 
	cin>>num>>changet;
	for(int i= 0; i < num; i++)
		cin>>catcht[i];

	for(int i= 0; i < num; i++)
	{
		for(int j= 0; j < num; j++)
		{
			timecheck[j].second= j;
			if(i == j)
				timecheck[j].first= catcht[j];
			else if(i > j)
				timecheck[j].first= catcht[j]+(i-j)*changet;
			else
				timecheck[j].first= catcht[j]+(num-j+i)*changet;
		}
 
		sort(timecheck, timecheck+num, SortComp);
		sum+= catcht[timecheck[0].second];

		if(i > timecheck[0].second)
			maxchange= max(maxchange, (long long int)i-timecheck[0].second);
		else if(i < timecheck[0].second)
			maxchange= max(maxchange, (long long int)num-timecheck[0].second+i(long long int));
	}
 
	sum+= changet*maxchange;
	cout<<sum<<endl;
 
	return 0;
}