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;
#pragma GCC optimize("Ofast")
void qread(int &x) {
  int neg = 1;
  x = 0;
  register char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = 10 * x + c - '0', c = getchar();
  x *= neg;
}
void Wl(int x) { printf("%d\n", x); }
void Wl(double x) { printf("%lf\n", x); }
void Wl(string x) { printf("%s\n", x.c_str()); }
void Wl(long long x) { printf("%lld\n", x); }
void W(int x) { printf("%d ", x); }
void W(double x) { printf("%lf ", x); }
void W(string x) { printf("%s ", x.c_str()); }
void W(long long x) { printf("%lld ", x); }
long long a[2005];
long long mn[2005];
int main() {
  int n;
  qread(n);
  ;
  int spell;
  qread(spell);
  ;
  long long ans = LLONG_MAX;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    mn[i] = a[i];
  }
  for (int use_spell = 0; use_spell < n; use_spell++) {
    long long cth = 0;
    for (int i = 0; i < n; i++) {
      cth += mn[i];
    }
    for (int i = 0; i < n; i++) {
      mn[i] = min(mn[i], a[((i - use_spell - 1) + n) % n]);
    }
    ans = min(ans, cth + use_spell * spell);
  }
  Wl(ans);
}
