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>
const int Mod = (int)1e9 + 7;
const int MX = 1073741822;
const long long MXLL = 9223372036854775807;
const int Sz = 1110111;
using namespace std;
inline void Read_rap() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
}
long long d[Sz];
int n;
long long x;
inline long long calc(long long a, long long b) {
  if (a < b) return (b - a) * x + d[a];
  return (n - a + b - 1) * x + d[a];
}
int main() {
  Read_rap();
  cin >> n >> x;
  for (int i = 1; i <= n; i++) cin >> d[i];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      d[i] = min(d[i], calc(j, i));
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      d[i] = min(d[i], calc(j, i));
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      d[i] = min(d[i], calc(j, i));
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += d[i];
  cout << ans;
  return 0;
}
