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;
signed main() {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (long long i = 0; i < (long long)(n); i++) {
    cin >> a[i];
  }
  static long long dp[2002][2002];
  for (long long i = 0; i < (long long)(2002); i++)
    for (long long j = 0; j < (long long)(2002); j++) dp[i][j] = (1LL << 59);
  for (long long i = 0; i < (long long)(n); i++) dp[i][0] = a[i];
  for (long long i = 0; i < (long long)(n); i++) {
    for (long long j = 1; j <= n; j++) {
      for (long long j = 0; j < (long long)(n + 1); j++) {
        dp[i][j] = min(dp[i][j - 1], a[(i - j + n) % n]);
      }
    }
  }
  long long ans = (1LL << 59);
  for (long long i = 0; i < (long long)(n + 1); i++) {
    long long tmp = 0;
    for (long long j = 0; j < (long long)(n); j++) {
      tmp += dp[j][i];
    }
    tmp += i * x;
    ans = min<long long>(ans, tmp);
  }
  cout << ans << endl;
}
