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
# pragma comment(linker, "/STACK: 2000000")
# include <stdio.h>
# include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
# define x first
# define y second
# define ll long long
# define IOS ios_base :: sync_with_stdio(0);cin.tie(0)
# define p(v) cerr << #v << " = " << v << '\n';
# define p2(v) cerr << #v << " = " << (complex < int > (v.x,v.y)) << '\n'
# define vi vector < int >
# define vll vector < ll >
# define pii pair < int , int >
int s[1 << 20];
int was[1 << 20];
int main(void)
{
    #ifdef CF
    freopen("input","r",stdin);
    #endif // CF
    srand(time(0));
    fo << fixed << setprecision(7);
    cerr << fixed << setprecision(7);
    int n,k;
    fi>>n>>k;
    for (int i = 1;i <= n;++i)
        fi>>s[i];
    ll ans = 0;
    auto get = [&](auto x,auto t)
    {
        return (x + t - 1) % n + 1;
    };
    ll cnt = 0;
    for (int t = 1;t <= n;++t)
    {
        pair < ll , int > mn = {(ll)(1e18),-1};
        for (int i = 1;i <= n;++i)
            if (!was[get(i,t)])
                mn = min(mn,{s[i] - s[get(i,t)],i});
        was[get(mn.y,t)] = 1;
        cnt += mn.x;
        cnt += k;
        ans = min(ans,cnt);
    }
    fo << (accumulate(s+1,s+1+n,0ll) + ans) << '\n';
    return 0;
}