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 int INF = (long long int)1e9 + 10;
const long long int INFLL = (long long int)1e18 + 10;
const long double EPS = 1e-12;
const long long int MOD = 1e9 + 7;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
template <class T>
inline T sq(T a) {
  return a * a;
}
long long int in() {
  long long int x;
  scanf("%lld", &x);
  return x;
}
char yuyushiki[1000010];
string stin() {
  scanf("%s", yuyushiki);
  return yuyushiki;
}
int n;
long long int x;
long long int a[2010];
int main() {
  n = in();
  x = in();
  for (long long int i = (0); i < (long long int)(n); i++) {
    a[i] = in();
  }
  long long int ans = INFLL;
  for (long long int i = (0); i < (long long int)(n); i++) {
    long long int res = 0;
    for (long long int j = (0); j < (long long int)(n); j++) {
      res += a[j];
    }
    chmin(ans, res + x * i);
    for (long long int j = (0); j < (long long int)(n); j++) {
      chmin(a[j], a[(j + n - 1) % n]);
    }
  }
  cout << ans << endl;
  return 0;
}
