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;
int main() {
  long long n, x;
  cin >> n >> x;
  long long a[2222];
  vector<pair<long long, long long> > vp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vp.push_back(make_pair(a[i], i));
  }
  sort(vp.begin(), vp.end());
  long long ans = 1LL << 60;
  for (int i = 0; i < n; i++) {
    long long ta[2222];
    for (int j = 0; j < n; j++) ta[j] = a[j];
    for (int j = 0; j < n; j++) {
      long long np = vp[j].second;
      long long cost = vp[j].first;
      if (ta[np] < cost) continue;
      for (int k = 0; k < i; k++) {
        if (ta[(np + k + 1) % n] < cost) break;
        ta[(np + k + 1) % n] = cost;
      }
    }
    long long sum = 0;
    for (int j = 0; j < n; j++) sum += ta[j];
    ans = min(ans, sum + x * i);
  }
  cout << ans << endl;
  return 0;
}
