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;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pii;
#define forr(i,a,n) for (int i = (a); i < (n); ++i)
#define rforr(i,a,n) for (int i = (a); i > (n); --i)
#define ALL(x) x.begin(),x.end()
#define LEN(X) ((int)(X).size())
#define CASET int _T; cin >> _T; forr(caset,0,_T)
#define int long long

main() {
    ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    
    int n,x; cin >> n >> x;
    int a[2001]; forr(i,0,n) cin >> a[i];

    int cost = 0;
    int offset[2001];
    forr(i,0,n) {
        int best = 0;
        forr(j,0,n) {
            int off = (i+j)%n;
            if (j*x+a[off] < best*x+a[(i+best)%n])
                best = j;
        }
        offset[i]=best;
    }

    int last = 0;
    forr(off,0,n) {
        bool found = false;
        forr(i,0,n) {
            if (offset[i]==off) {
                cost += a[(off+i)%n];
                found = true;
            }
        }
        if (found) {
            cost += (off-last)*x;
            last = off;
        }
    }
    cout << cost << '\n';


        
}