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 rep(i,n) for(int i=0;i<(int)n;++i)
#define rrep(i,n) for(int i=(int)n-1;i>=0;--i)
using namespace std;
using ll = long long;
template<typename T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T> inline bool chmin(T& a,T b){if(b<a){a=b;return 1;}return 0;}
template<typename T> vector<T> make_vec(size_t a){return vector<T>(a);}
template<typename T,typename... Ts>
auto make_vec(size_t a,Ts... ts){return vector<decltype(make_vec<T>(ts...))>(a,make_vec<T>(ts...));}
template<typename T,typename U,typename... V>
typename enable_if<is_same<T,U>::value>::type fill_v(U& u,const V... v){u=U(v...);}
template<typename T,typename U,typename... V>
typename enable_if<!is_same<T,U>::value>::type fill_v(U& u,const V... v){for(auto& e:u)fill_v<T>(e,v...);}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

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

    ll ans = 0;
    ll magic = 0;
    const ll INF = 1e18;
    rep(i, n) {
        ll now = INF;
        ll now_magic = 0;
        for (int j = n; j >= 1; --j) {
            if (chmin(now, (n-j) * x + a[(i+j) % n])) chmax(now_magic, n-j);
        }
        ans += now - now_magic * x;
        chmax(magic, now_magic);
    }
    ans += magic * x;
    cout << ans << endl;
}
