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;
long double TL = 1.0;
bool is_tl = 0;
long long CALLS = 0;
inline bool IS() {
  if (++CALLS == 1000) {
    CALLS = 0;
    is_tl |= clock() > (TL - 0.1) * CLOCKS_PER_SEC;
  }
  return is_tl;
}
template <typename T1, typename T2>
inline void amin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void amax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  os << p.first << ' ' << p.second << endl;
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  for (auto &u : v) os << u << ' ';
  os << endl;
  return os;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 2e5 + 7, NS = 2e3 + 7, lg = 19, sq = 550, inf = 1e18 + 7,
                SM = 1e2 + 7;
const long double eps = 1e-9;
long long hash_mod[4] = {1000000007, 998244353, 1000000009, 999999937},
          mod = hash_mod[rnd() % 4];
long long hash_pows[4] = {179, 239, 1007, 2003}, P = hash_pows[rnd() % 4];
void solve() {
  long long n, x, ans = inf;
  cin >> n >> x;
  vector<long long> a(n), b;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) {
    long long cur = 0;
    for (long long j = 0; j < n; j++) {
      amin(a[j], a[(j - i + n) % n]);
      cur += a[j];
    }
    amin(ans, cur + x * i);
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
