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;
const long long mod = 1e9 + 7;
const int infi = 2147483600;
const long long infl = 1e17;
int 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];
  vector<long long> MIN(n, infl);
  long long ans = 0;
  for (long long i = 0; i < (long long)n; i++) {
    long long now = infl;
    for (long long j = 0; j < (long long)n; j++) {
      if (i >= j) {
        now = min(now, x * (i - j) + a[j]);
      } else {
        now = min(now, x * (n + i - j) + a[j]);
      }
    }
    MIN[i] = now;
  }
  for (long long i = 0; i < (long long)n; i++) {
    if (MIN[i] == a[i]) {
      for (long long j = (long long)1; j < (long long)n + 1; j++) {
        if (MIN[(i + j) % n] != a[i] + x * j) {
          ans += a[i] * j + x * (j - 1);
          break;
        }
      }
    }
  }
  cout << ans << endl;
}
