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 MAXN = 4005;
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) {
    putchar('-');
    x = (~x) + 1;
  }
  if (9 < x) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename _T>
_T MIN(const _T a, const _T b) {
  return a < b ? a : b;
}
int q[MAXN];
int a[MAXN];
int N, x;
long long calc(const int l) {
  long long ret = l * x;
  int h = 1, t = 0, r = 1;
  for (int i = 1; i <= N; i++) {
    while (h <= t && q[h] < i) h++;
    for (; r <= i + l; r++) {
      while (h <= t && a[q[t]] >= a[r]) t--;
      q[++t] = r;
    }
    ret += a[q[h]];
  }
  return ret;
}
int main() {
  read(N), read(x);
  for (int i = 1; i <= N; i++) read(a[i]), a[i + N] = a[i];
  long long ans = 1e18;
  for (int i = 0; i < N; i++) ans = MIN(ans, calc(i));
  write(ans), putchar('\n');
  return 0;
}
