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 N, x;
long long inc(long long i) {
  if (i == N - 1)
    return 0;
  else
    return i + 1;
}
long long dec(long long i) {
  if (i == 0)
    return N - 1;
  else
    return i - 1;
}
int main() {
  scanf("%d %d", &N, &x);
  vector<long long> a(N);
  long long m = 999999999, index = 0;
  ;
  for (long long i = 0; i < (N); i++) {
    scanf("%d", &a[i]);
    if (m > a[i]) {
      m = a[i];
      index = i;
    }
  }
  vector<long long> p(N), call(N), num(N);
  for (long long i = 0; i < (N); i++) {
    call[i] = 0;
    num[i] = 0;
  }
  p[index] = a[index];
  call[index] = 1;
  long long i = inc(index);
  while (1) {
    if (i == index) break;
    if (p[dec(i)] + x < a[i]) {
      num[i] = num[dec(i)] + 1;
      p[i] = p[dec(i)] + x;
      long long j = i;
      while (1) {
        j = dec(j);
        if (call[j] != 0) {
          call[j]++;
          break;
        }
      }
    } else {
      call[i]++;
      p[i] = a[i];
    }
    i = inc(i);
  }
  long long ans = 0;
  long long M = 0;
  for (long long i = 0; i < (N); i++) {
    ans += call[i] * a[i];
    M = max(num[i], M);
  }
  ans += M * x;
  printf("%lld\n", ans);
}
