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;
bool SortComp(pair<long long int, int> p1, pair<long long int, int> p2) {
  if (p1.first < p2.first)
    return true;
  else
    return false;
}
int main(void) {
  int num;
  long long int changet, catcht[2000] = {}, maxchange = 0, sum = 0;
  pair<long long int, int> timecheck[2000];
  cin >> num >> changet;
  for (int i = 0; i < num; i++) cin >> catcht[i];
  for (int i = 0; i < num; i++) {
    for (int j = 0; j < num; j++) {
      timecheck[j].second = j;
      if (i == j)
        timecheck[j].first = catcht[j];
      else if (i > j)
        timecheck[j].first = catcht[j] + (i - j) * changet;
      else
        timecheck[j].first = catcht[j] + (num - j + i) * changet;
    }
    sort(timecheck, timecheck + num, SortComp);
    sum += catcht[timecheck[0].second];
    if (i == timecheck[0].second)
      maxchange = max(maxchange, (long long int)0);
    else if (i > timecheck[0].second)
      maxchange = max(maxchange, (long long int)(i - timecheck[0].second));
    else
      maxchange =
          max(maxchange, (long long int)(num - timecheck[0].second + i));
  }
  sum += changet * maxchange;
  cout << sum << endl;
  return 0;
}
