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"
#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a),i##_len=(b);i<i##_len;++i)
#define MSVC_UNKO(x) x
#define rep(...) MSVC_UNKO(_overload3(__VA_ARGS__,repi,_rep,_rep)(__VA_ARGS__))
#define all(c) c.begin(),c.end()
#define mp make_pair
#define write(x) cout<<(x)<<"\n"
using namespace std; typedef long long ll; template<class T>using vv = vector<vector<T>>;
template<class T>auto vvec(int n, int m, T v) { return vv<T>(n, vector<T>(m, v)); }
template<class T1, class T2>bool chmax(T1& a, const T2& b) { return a < b ? a = b, 1 : 0; }
template<class T1, class T2>bool chmin(T1& a, const T2& b) { return b < a ? a = b, 1 : 0; }
constexpr int INF = 1 << 29, MOD = int(1e9) + 7; constexpr ll LINF = 1LL << 60;
struct aaa { aaa() { cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(10); }; }aaaa;

int main() {
    ll N, x;
    cin >> N >> x;
    vector<int> a(N);
    rep(i, N) cin >> a[i];

    ll ans = LINF;
    for (int d = 1; d <= N; ++d) {
        for (int i = 0; i < d; ++i) {
            ll t = 0;
            for (int s = 0; s < N; s += d) {
                t += a[(i + s) % N];
            }
            ans = min(ans, t * d + x * (d - 1));
        }
    }

    write(ans);
}