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 ld = long double;
using P = pair<int, int>;
using Graph = vector<vector<int>>;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int main() {
  ll N, x;
  cin >> N >> x;
  vector<ll> a(N);
  vector<vector<pair<ll, ll>>> table(N, vector<pair<ll, ll>>(N));
  for (int i = 0; i < N; i++) cin >> a[i];
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (j >= i) {
        ll cost, cnt;
        cnt = j - i;
        cost = a[i] + cnt * x;
        table[i][j] = {cost, cnt};
      } else {
        ll cost, cnt;
        cnt = N - i + j;
        cost = a[i] + cnt * x;
        table[i][j] = {cost, cnt};
      }
    }
  }
  ll ans = 0;
  ll max_cnt = 0;
  for (int j = 0; j < N; j++) {
    ll cost_min = 100000000000;
    ll cnt = 0;
    for (int i = 0; i < N; i++) {
      if (cost_min > table[i][j].first) {
        cost_min = table[i][j].first;
        cnt = table[i][j].second;
      }
    }
    ans += cost_min - cnt * x;
    max_cnt = max(max_cnt, cnt);
  }
  ans += max_cnt * x;
  cout << ans << endl;
}
