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;
template <class T, class U>
using P = pair<T, U>;
template <class T>
using Heap = priority_queue<T>;
template <class T>
using heaP = priority_queue<T, vector<T>, greater<T>>;
template <class T>
bool ChangeMax(T& a, const T& b) {
  if (a >= b) return false;
  a = b;
  return true;
}
template <class T>
bool ChangeMin(T& a, const T& b) {
  if (a <= b) return false;
  a = b;
  return true;
}
template <class T, size_t N, class U>
void Fill(T (&a)[N], const U& v) {
  fill((U*)a, (U*)(a + N), v);
}
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& e : v) is >> e;
  return is;
}
int main() {
  ll n, x;
  cin >> n >> x;
  vector<ll> a(n);
  cin >> a;
  vector<ll> v(n, 1001001001);
  ll ans = 1001001001;
  auto f = [&](ll k) -> ll {
    ll res = k * x;
    for (ll i = (ll)(0); i < (ll)(n); ++i) {
      ChangeMin(v[i], a[(i - k + n) % n]);
      res += v[i];
    }
    return res;
  };
  for (ll i = (ll)(0); i < (ll)(n); ++i) ChangeMin(ans, f(i));
  cout << ans << endl;
  return 0;
}
