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;
using ll = long long;
using VI = vector<int>;
using VL = vector<ll>;
constexpr ll INF = 1000000000;
constexpr ll LLINF = 1LL << 60;
constexpr ll mod = 1000000007;
constexpr ll MOD = 998244353;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
inline void init() {
  cout.tie(nullptr);
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
}
int main() {
  init();
  int n;
  ll x;
  cin >> n >> x;
  VL a(n, INF);
  ll ans = 0;
  for (int(i) = (0); (i) < (n); (i)++) {
    if (!i) cin >> a[i];
    ans += a[i];
  }
  vector<VL> Min(n, VL(n));
  Min[0] = a;
  for (int(i) = (1); (i) < (n); (i)++) {
    ll sum = 0;
    for (int(j) = (0); (j) < (n); (j)++) {
      Min[i][j] = min(Min[i - 1][j], a[(j - i + n) % n]);
      sum += Min[i][j];
    }
    chmin(ans, sum + x * i);
  }
  cout << ans << "\n";
}
