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 N;
long long x, A[4000];
vector<pair<long long, int> > B;
bool flg[2000] = {};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(9);
  cin >> N >> x;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    A[N + i] = A[i];
    B.push_back(make_pair(A[i], i));
  }
  sort(B.begin(), B.end());
  long long ans = (long long)9223372036854775807;
  for (int i = 0; i < N; i++) {
    flg[B[i].second] = true;
    int s;
    for (int j = 0; j < N; j++) {
      if (flg[j]) {
        s = j;
        break;
      }
    }
    long long tmp = 0;
    long long bef = 0;
    long long maxX = 0;
    long long nowX = 0;
    for (int j = 0; j < N; j++) {
      if (flg[j + s]) {
        tmp += A[j + s];
        bef = A[j + s];
        nowX = 0;
      } else {
        tmp += bef;
        nowX++;
        maxX = max(maxX, nowX);
      }
    }
    ans = min(ans, tmp + x * maxX);
  }
  cout << ans << endl;
  return 0;
}
