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;
char s[100006];
struct arr {
  int x, i;
} a[100006];
bool operator<(arr x, arr y) { return x.x < y.x; }
long long res, ans;
int n, m, k, x, y, l, r, cnt, f[100006], p[100006];
inline int read() {
  int x = 0, c = getchar();
  while (c < 48) c = getchar();
  while (c > 47) x = x * 10 + c - 48, c = getchar();
  return x;
}
void print(long long x) {
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
int main() {
  scanf("%d%d", &n, &x);
  ans = 1e18;
  for (int i = 1; i <= n; i++) a[i].x = read(), a[i].i = i;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    f[a[i].i] = 1;
    int last = 0, first = 0, tmp = 0;
    long long res = 0;
    for (int i = 1; i <= n; i++)
      if (f[i]) {
        if (last)
          tmp = max(tmp, i - last - 1), res += 1LL * (i - last) * a[last].x;
        else
          first = i;
        last = i;
      }
    tmp = max(tmp, first - 1 + n - last);
    res += 1LL * (first + n - last) * a[last].x;
    ans = min(ans, 1LL * tmp * x + res);
  }
  printf("%lld\n", ans);
}
