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>
int a[2005];
int l, m, n, r, x, y;
long long s[2005];
long long sm;
int sub(int p, int q) {
  if (q == 0) {
    m = 1000000001;
    for (int k = 1 - p; k <= 0; k++) {
      if (k < 0)
        l = k + n;
      else
        l = k;
      if (a[l] <= m) {
        m = a[l];
        y = l;
      }
    }
    return m;
  } else {
    if (q - p < 0)
      r = q - p + n;
    else
      r = q - p;
    if (y != r) {
      if (a[y] < a[q])
        return a[y];
      else {
        y = q;
        return a[q];
      }
    } else {
      m = 1000000001;
      for (int k = q - p + 1; k <= q; k++) {
        if (k < 0)
          l = k + n;
        else
          l = k;
        if (a[l] <= m) {
          m = a[l];
          y = l;
        }
      }
      return m;
    }
  }
}
int main() {
  for (int i = 0; i < 2005; i++) {
    a[i] = 0;
    s[i] = 0;
  }
  sm = 2000000000001;
  scanf("%d %d", &n, &x);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; i++) {
    s[i] = (i - 1) * x;
    for (int j = 0; j < n; j++) {
      m = sub(i, j);
      s[i] = s[i] + m;
    }
    if (sm > s[i]) sm = s[i];
  }
  printf("%lld", sm);
  return 0;
}
