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<bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#include "../../dump.hpp"
#else
#define dump(...)
#endif

#define rep(i,n) for(int i=0,i##_cond=(n);i<i##_cond;i++)
#define FOR(i,a,b) for(int i=(a),i##_cond=(b);i<i##_cond;i++)
#define ROF(i,a,b) for(int i=(a)-1,i##_cond=(b);i>=i##_cond;i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend() //sortで大きい順
#define UNIQUE(v) v.erase(unique(all(v)),v.end())
#define SUM(a) accumulate(all(a),0)
#define sz(x) ((int)(x).size())
#define pb push_back
#define fst first
#define snd second
#define mp make_pair

typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<ll,ll> pii;

#define int ll

const int inf = 1ll<<62;
const int mod = 1e9+7;

main(){
  int n,x; cin >> n >> x;
  vi a(n); rep(i,n) cin >> a[i];
  vi b(n); rep(i,n) b[i] = a[i];

  int ans = SUM(a);
  rep(i,n-1){
    int tmp = 0;
    rep(j,n){
      b[j] = min(b[j], a[(j-i-1+n)%n]);
      tmp = (i+1)*x+SUM(b);
    }
    ans = min(ans, tmp);
  }
  cout << ans << endl;
}