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>
long long int shi[2005];
long long int you[2005];
long long int duiz[2005];
long long int xuan[2005];
int main() {
  long long int a, x;
  while (~scanf("%lld %lld", &a, &x)) {
    long long int shou;
    long long int wei = 0;
    for (int i = 0; i < a; i++) {
      scanf("%lld", &shi[i]);
      if (i == 0)
        shou = shi[i];
      else if (shou > shi[i]) {
        shou = shi[i];
        wei = i;
      }
    }
    int dui = 0;
    duiz[dui] = shou;
    you[dui] = 1;
    for (int i = 0; i < a - 1; i++) {
      wei++;
      if (wei == a) wei = 0;
      if (shi[wei] < shou + x) {
        shou = shi[i];
        shou = shi[wei];
        dui++;
        duiz[dui] = shou;
        you[dui] = 1;
      } else {
        you[dui]++;
      }
    }
    xuan[dui] = (duiz[dui] + x) * you[dui] - x;
    long long int xian = you[dui];
    long long int kais;
    for (int i = dui - 1; i >= 0; i--) {
      xian += you[i];
      kais = you[dui];
      xuan[i] = (duiz[i] + x) * xian - x;
      for (int j = i + 1; j <= dui; j++) {
        if (kais * (duiz[i] + x) - x + xuan[j] < xuan[i])
          xuan[i] = kais * (duiz[i] + x) - x + xuan[j];
        kais += you[j];
      }
    }
    printf("%lld\n", xuan[0]);
  }
  return 0;
}
