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;
ll INF = 1e9 + 7;
int main() {
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<ll> tmp = a;
  ll chmax = 0;
  ll chpos = 0;
  set<ll> s;
  set<ll> ch;
  for (ll i = 0; i < n; i++) {
    s.insert(i);
  }
  vector<ll> org(n);
  for (ll i = 0; i < n; i++) {
    ll mi = INF;
    ll pos = 0;
    for (ll j = 0; j < n; j++) {
      if (s.count(j) == 0) continue;
      if (tmp[j] < mi) {
        mi = tmp[j];
        pos = j;
      }
    }
    s.erase(pos);
    for (ll j = 0; j < pos; j++) {
      if (s.count(j) == 0) continue;
      if (tmp[j] > x * (n - pos + j) + mi) {
        ch.insert(j);
        tmp[j] = x * (n - pos + j) + mi;
        org[j] = pos;
        if (chmax < n - pos + j) {
          chmax = n - pos + j;
          chpos = j;
        }
      }
    }
    for (ll j = pos + 1; j < n; j++) {
      if (s.count(j) == 0) continue;
      if (tmp[j] > x * (j - pos) + mi) {
        ch.insert(j);
        tmp[j] = x * (j - pos) + mi;
        org[j] = pos;
        if (chmax < j - pos) {
          chmax = j - pos;
          chpos = j;
        }
      }
    }
  }
  ll ans = 0;
  for (ll i = 0; i < n; i++) {
    if (ch.count(i) == 0)
      ans += a[i];
    else
      ans += a[org[i]];
  }
  ans += chmax * x;
  cout << ans << endl;
}
