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;
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
const double PI = acos(-1.0);
const long long mod = 1000000007;
const long long INF = 1e9;
const long long MAXN = 100001;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = (0); i < (n); ++i) cin >> a[i];
  long long ans = 0;
  vector<pair<long long, long long> > b(n);
  for (int i = (0); i < (n); ++i) b[i] = make_pair(a[i], i);
  sort(b.begin(), b.end());
  vector<char> ye(n, 0);
  vector<char> ny(n, 0);
  bool ok = 0;
  while (1) {
    int ind = -1;
    for (int i = (0); i < (n); ++i)
      if (ye[b[i].second] == 0) {
        ind = i;
        break;
      }
    if (ind == -1) break;
    if (x >= b[ind].first || !ok) {
      ye[b[ind].second] = 1;
      ok = 1;
      ans += b[ind].first;
    } else {
      ans += x;
      ny.assign(n, 0);
      for (int i = (0); i < (n); ++i)
        if (ye[i]) ny[(i + 1) % n] = 1;
      ye = ny;
    }
  }
  cout << ans;
  return 0;
}
