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>
template <class T>
inline void rd(T &x) {
  char c = getchar();
  x = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
using namespace std;
const int mod = 1e9 + 7;
const int N = 2e3 + 10;
int n, x, a[N], mi[N][N];
long long ans;
int main() {
  rd(n), rd(x);
  for (int i = 1; i <= n; i++) rd(a[i]);
  for (int i = 1; i <= n; i++) {
    mi[i][i] = a[i];
    ans += a[i];
    for (int j = i + 1; j <= n; j++) mi[i][j] = min(a[j], mi[i][j - 1]);
  }
  for (int l = 1; l < n; l++) {
    long long now = l * x;
    for (int i = 1; i <= n; i++) {
      if (i > l)
        now += mi[i - l][i];
      else
        now += min(mi[1][i], mi[i + n - l][n]);
    }
    ans = min(ans, now);
  }
  printf("%lld\n", ans);
  return 0;
}
