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;
long long colors[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, x;
  cin >> n >> x;
  for (long long i = 0; i < n; ++i) {
    cin >> colors[i];
  }
  long long sol = 0, mincost, mincol, maxdist = 0;
  for (long long i = 0; i < n; ++i) {
    mincost = colors[i];
    mincol = i;
    for (long long j = 0; j < n; ++j) {
      if (j <= i && mincost > colors[j] + x * (i - j)) {
        mincost = colors[j] + x * (i - j);
        mincol = j;
      } else if (j > i && mincost > colors[j] + x * (n + i - j)) {
        mincost = colors[j] + x * (n + i - j);
        mincol = j;
      } else if (j <= i && mincost == colors[j] + x * (i - j)) {
        if (mincol <= i && (i - j) < (i - mincol)) {
          mincol = j;
        } else if (mincol > i && (i - j) < (n + i - mincol)) {
          mincol = j;
        }
      } else if (j > i && mincost == colors[j] + x * (n + i - j)) {
        if (mincol <= i && (n + i - j) < (i - mincol)) {
          mincol = j;
        } else if (mincol > i && (n + i - j) < (n + i - mincol)) {
          mincol = j;
        }
      }
    }
    sol += colors[mincol];
    if (mincol <= i) {
      maxdist = max(maxdist, x * (i - mincol));
    } else {
      maxdist = max(maxdist, x * (n + i - mincol));
    }
  }
  sol += maxdist;
  cout << sol << "\n";
  return 0;
}
