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;
const int IINF = 1e9 + 10;
const long long LLINF = (long long)1e18 + 10;
const long long MOD = (long long)1e9 + 7;
const long long inv = MOD - 2;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1},
          dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};
long long n, x;
vector<long long> a;
pair<long long, long long> calc(int i, int j) {
  pair<long long, long long> res;
  if (i < j) {
    res.first = a[j] + (n - j + i) * x;
    res.second = n - j + i;
  } else {
    res.first = a[j] + (i - j) * x;
    res.second = i - j;
  }
  return res;
}
int main() {
  cin >> n >> x;
  a.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long magic = 0;
  vector<long long> cnt(n);
  for (long long i = 0; i < n; i++) {
    long long cost = a[i], cap = i, mgccnt = 0;
    for (long long j = 0; j < n; j++) {
      pair<long long, long long> p = calc(i, j);
      if (p.first < cost) {
        cost = p.first;
        cap = j;
        mgccnt = p.second;
      }
    }
    cnt[cap]++;
    magic = max(magic, mgccnt);
  }
  long long ans = magic * x;
  for (long long i = 0; i < n; i++) {
    ans += cnt[i] * a[i];
  }
  cout << ans << endl;
  return 0;
}
