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 <stdio.h>
#include <bits/stdc++.h>			

#define pb push_back
#define pf push_front
#define pp pop_back
#define sz(a) (int)(a.size())
#define mp make_pair
#define F first
#define S second
#define next _next
#define prev _prev
#define left _left
#define right _right
#define y1 _y1
#define all(x) x.begin(), x.end()
#define what_is(x) #x << " is " << (x)
#define int ll

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int N = (int)1e6 + 123;
const ll INF = (ll)1e18 + 123;
const int inf = (int)1e9 + 123;
const int MOD = (int)1e9 + 7;

int n, x, a[4000];
int opt[4000];
ll dp[4000];

main() {
	cin >> n >> x;
	int mn = inf, mn_pos = -1;
	for(int i = 1; i <= n; i ++) {
		cin >> a[i];
		if(a[i] < mn) mn = a[i], mn_pos = i;		
	}
	ll res = INF;
	for(int i = n + 1; i <= 2 * n; i ++)
		a[i] = a[i - n];
	int itr = 0;
	for(int i = 1; i <= n; i ++) {
		if(itr == 100)	break;
		if(a[i] == mn) {
			itr ++;
			dp[i - 1] = 0;
			for(int j = i; j <= i + n - 1; j ++) {
				dp[j] = INF;
				for(int k = j; k >= i; k --) {
					if(dp[j] > dp[k - 1] + 1ll * a[k] * (j - k + 1) + 1ll * x * (j - k))
						dp[j] = dp[k - 1] + 1ll * a[k] * (j - k + 1) + 1ll * x * (j - k);
					//else break;
				}
			}
			res = min(res, dp[i + n - 1]);
		}
	}
	cout << res;
	return 0;  
}
