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;
using ll = long long;
using db = double;
const ll mod = 1000000007;
const ll INF = 1000000099;
vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
signed main() {
  ll n, x, ans = 0, mxzrs = -1;
  cin >> n >> x;
  vector<ll> v(2010, INF);
  for (int i = 0; i < n; i++) {
    cin >> v[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    ll mnind = 0, mncost = 5 * INF * INF;
    for (int j = 1; j <= n; j++) {
      if (i >= j && mncost > v[j] + x * ((i - j) % n)) {
        mnind = j;
        mncost = v[j] + x * ((i - j) % n);
      } else if (mncost > v[j] + x * ((n - (j - i)) % n)) {
        mnind = j;
        mncost = v[j] + x * ((n - (j - i)) % n);
      } else if (i >= j && mncost == v[j] + x * ((i - j) % n) &&
                 mxzrs >= ((i - j) % n)) {
        mnind = j;
        mncost = v[j] + x * ((i - j) % n);
      } else if (mncost == v[j] + x * ((n - (j - i)) % n) &&
                 mxzrs >= ((n - (j - i)) % n)) {
        mnind = j;
        mncost = v[j] + x * ((n - (j - i)) % n);
      }
    }
    ans += v[mnind];
    if (i >= mnind)
      mxzrs = max(mxzrs, (i - mnind) % n);
    else
      mxzrs = max(mxzrs, (n - (mnind - i)) % n);
  }
  ans += x * mxzrs;
  cout << ans << endl;
}
