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;
const ll inf = (1e18);
const ll mod = 1000000007;
const ll mod2 = 998244353;
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll c, ll d) { return c / gcd(c, d) * d; }
vector<vector<pair<ll, ll>>> slimes;
int main() {
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  slimes.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < n; j++) {
      slimes[(i + j) % n].push_back(pair<ll, ll>(a[i] + j * x, j));
    }
  }
  ll ans = 0, change = 0, need_c = 0;
  for (int i = 0; i < n; i++) {
    sort((slimes[i]).begin(), slimes[i].end());
    ans += slimes[i][0].first;
    change += slimes[i][0].second;
    need_c = max(need_c, slimes[i][0].second);
  }
  ans -= x * (change - need_c);
  cout << ans << endl;
}
