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 ll = long long;
using namespace std;
struct RMQ {
  const ll INF = 1ll << 60;
  int n;
  vector<ll> data;
  RMQ() {}
  RMQ(int n_) { init(n_); }
  RMQ(int n_, vector<ll> &vec) {
    init(n_);
    set(vec);
    build();
  }
  void init(int n_) {
    n = 1;
    while (n < n_) n <<= 1;
    data.assign(n << 1, INF);
  }
  void set(vector<ll> &vec) {
    for (int i = 0; i < (int)vec.size(); i++) data[i + n] = vec[i];
  }
  void build() {
    for (int i = n - 1; i > 0; i--)
      data[i] = min(data[i << 1], data[(i << 1) | 1]);
  }
  void update(int i, ll x) {
    i += n;
    data[i] = x;
    while (i >>= 1) data[i] = min(data[i << 1], data[(i << 1) | 1]);
  }
  ll query(int a, int b) {
    ll res = INF;
    for (int l = a + n, r = b + n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) res = min(res, data[l++]);
      if (r & 1) res = min(res, data[--r]);
    }
    return res;
  }
};
int main() {
  int n, x;
  cin >> n >> x;
  vector<ll> a(n << 1);
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    a[i + n] = a[i];
  }
  RMQ data(n << 1, a);
  ll ans = 1ll << 60;
  for (int i = 0; i < (n); i++) {
    ll ch = i * x;
    for (int j = 0; j < (n); j++) ch += data.query(n + j - i, n + j + 1);
    if (ch < ans) ans = ch;
    cout << ch << endl;
  }
  cout << ans << endl;
  return 0;
}
