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<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<algorithm>

#define INT unsigned long long int
#define FOR1(n) for(int i=0;i<n;i++)
#define FOR2(j, n) for(int j=0;j<n;j++)
#define FOR3(i, m, n) for(int i=m;i<=n;i++)
#define PINT(d) printf("%d\n", d);
#define PDOUBLE(lf) printf("%lf\n", lf);
#define PSTR(s) printf("%s\n", s);
#define ALL(c) c.begin(), c.end()
#define MOD 1000000007
#define MAX_N 100000

using namespace std;

main(){
	long long int n, x;
	cin >> n >> x;
	long long int a[n];
	FOR1(n) cin >> a[i];
	long long int b[n][n], max_x_count = 0, x_count[n] = {};
	long long int ans = 0;
	
	//魔法を唱える回数をk回に固定する
	FOR2(k, n){
		FOR2(i, n){
			long long int tmp;
			if(i - k >= 0){
				tmp = a[i-k];
			}else{
				tmp = a[n - k + i];
			}

			if(k == 0){
				b[k][i] = tmp;
			}else{
				if(tmp + k * x < b[k-1][i] + x_count[i] * x){
					b[k][i] = tmp;
					max_x_count = k;
					x_count[i] = k;
				}else{
					b[k][i] = b[k-1][i];
				}
			}
		}
	}
	
	FOR1(n){
		ans += b[n-1][i];
	}
	ans += x * max_x_count;
	
	cout << ans;
}