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>
#pragma GCC optimize("Ofast,fast-math,unroll-loops")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<ll, ll>;
using pld = pair<ld, ld>;
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (auto &i : a) in >> i;
  return in;
}
template <class T>
ostream &operator<<(ostream &out, vector<T> a) {
  for (auto &i : a) out << i << ' ';
  return out;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, pair<T, U> p) {
  out << p.first << " " << p.second << " ";
  return out;
}
inline void Start() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
signed main() {
  Start();
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  cin >> a;
  vector<ll> b(n);
  for (ll i = 0; i < n; ++i) {
    ll cur = a[i];
    for (ll j = 0; j < n; ++j) cur = min(cur, a[(i - j + n) % n]), b[j] += cur;
  }
  ll ans = 1e9;
  for (ll i = 0; i < n; ++i) ans = min(ans, i * x + b[i]);
  cout << ans;
  return 0;
}
