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 vvi = vector<vi>;
using vl = vector<ll>;
using pi = pair<int, int>;
using vpi = vector<pi>;
using pl = pair<ll, ll>;
using vpl = vector<pl>;
template <class T>
bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  } else
    return false;
}
template <class T>
bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  } else
    return false;
}
template <class T>
void print(std::vector<T> a) {
  int sz = a.size();
  for (int i = 0; i < a.size(); i++) cout << a[i] << (i < sz - 1 ? ' ' : '\n');
}
void ioboost() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
}
int main(void) {
  freopen("input.txt", "r", stdin);
  ll n, x;
  cin >> n >> x;
  vl a(n), b(n);
  for (size_t i = 0; i < n; i++) {
    cin >> a[i];
  }
  ll ans = 1e18;
  for (int k = 0; k < n; k++) {
    ll cur = 0;
    for (int i = 0; i < n; i++) {
      if (k == 0) {
        b[i] = a[i];
      } else {
        chmin(b[i], a[(i - k + n) % n]);
      }
      cur += b[i];
    }
    chmin(ans, cur + (ll)k * x);
  }
  cout << ans << '\n';
}
