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;
void swap(long long int *a, long long int *b) {
  long long int tmp = *a;
  *a = *b;
  *b = tmp;
}
void babbleSort(long long int *a, long long int num) {
  for (int i = 0; i < num; i++) {
    for (int j = num - 1; j > i; j--) {
      if (a[j] >= a[j - 1]) {
        swap(&a[j], &a[j - 1]);
      }
    }
  }
}
long long int min(long long int a, long long int b) {
  if (a >= b)
    return b;
  else
    return a;
}
long long int max(long long int a, long long int b) {
  if (a >= b)
    return a;
  else
    return b;
}
int main() {
  long long int n, x, *a, cnt = 0, *from, *magicCnt;
  cin >> n >> x;
  a = new long long int[n];
  from = new long long int[n];
  magicCnt = new long long int[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    from[i] = i;
    magicCnt[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 1; j < n; j++) {
      if (a[(i + j) % n] > a[i] + x * j) {
        from[(i + j) % n] = i;
        magicCnt[(i + j) % n] = magicCnt[i] + j;
        a[(i + j) % n] = a[i] + x * j;
      }
    }
  }
  long long int ans = 0;
  long long int maxMagicCnt = 0;
  for (int i = 0; i < n; i++) {
    ans += a[from[i]];
    maxMagicCnt = max(magicCnt[i], maxMagicCnt);
  }
  ans += maxMagicCnt * x;
  cout << ans << endl;
  return 0;
}
