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;
unsigned long long min(unsigned long long a, unsigned long long b) {
  return (a > b) ? b : a;
}
unsigned long long keiro(unsigned long long n, unsigned long long k,
                         unsigned long long i) {
  if (k < i) {
    return (n - i) + k;
  } else {
    return k - i;
  }
}
int main(void) {
  int n, x;
  unsigned long long h = 0;
  cin >> n >> x;
  vector<int> a(n);
  vector<unsigned long long> ans(n);
  vector<unsigned long long> ans1(n);
  unsigned long long res = 0;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int k = 0; k < n; k++) {
    unsigned long long tmp = a[k];
    ans1[k] = k;
    for (int i = 0; i < n; i++) {
      if (a[i] + keiro(n, k, i) * x < tmp) {
        if (keiro(n, k, i) > ans[k]) {
          ans[k] = keiro(n, k, i);
        }
        tmp = a[i] + keiro(n, k, i) * x;
        ans1[k] = i;
      }
    }
  }
  h = *max_element(ans.begin(), ans.end());
  for (int k = 0; k < n; k++) {
    res += a[ans1[k]];
  }
  cout << h * x + res << endl;
}
