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>
const int INF = 1e9;
const int MOD = 1e9 + 7;
const long long LINF = 1e18;
using namespace std;
void CINT() {}
template <class Head, class... Tail>
void CINT(Head&& head, Tail&&... tail) {
  cin >> head;
  CINT(move(tail)...);
}
int main() {
  int n, x;
  CINT(n, x);
  vector<long long> a(n);
  for (auto& youso_ : (a)) cin >> youso_;
  vector<vector<long long> > vec(n, vector<long long>(n));
  for (int i = 0; i < (n); ++i) vec[i][0] = a[i];
  for (int j = (1); j < (n); ++j)
    for (int i = 0; i < (n); ++i)
      vec[i][j] = min(vec[i][j - 1], vec[(i - 1 + n) % n][j - 1]);
  long long ans = LINF;
  for (int bound = 0; bound < (n); ++bound) {
    long long cou = 0;
    for (int i = 0; i < (n); ++i) {
      cou += vec[i][bound];
    }
    ans = min(ans, cou + bound * x);
  }
  cout << (ans) << endl;
  return 0;
}
