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
v/* it was worth becoming a chemist */
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef double db;
typedef long double ld;
typedef unsigned int uni;
typedef unsigned long long unll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef vector<int> vi;
typedef vector<long long> vll;

#define mp make_pair
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)x.size()
#define NAME "puts"
#define F first
#define S second

const ll INF = 1e18;
const int inf = 1e9;
const int mod = 1e9 + 7;
const db EPS = (db) 1e-9;
const db pi = acos(-1.0);
const int MAXN = 1e5 + 5;

ll a, b, c;

int main() {
  	

    cin >> a >> b >> c;
    if (a % 2 == 0 || b % 2 == 0 || c % 2 == 0) {
    	cout << 0;
    	return 0;
    }
    cout << min(a, min(b, c)) * ((a + b + c) - min(a, min(b, c)) - max(a, max(b, c)));    
    return 0;
}
