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 <iostream>
#include <vector>
#include <algorithm>
#include <array>
using namespace std;
using li = long;
li N, X;
array<li, 2000> A;
vector<li>B;


void solve(){
  long long ans = 10000000000000;
  for (li i = 0; i < N; i++){
    for (li j = 0; j < N; j++){
      if(i > j){
        if (B[j] > A[j + N- i]) B[j] = A[j + N - i];
      }
      else{
        if(B[j] > A[j - i]) B[j] = A[j - i];
      }
    }
    long long tmp = accumulate(B.begin(), B.end(), 0) + i * X;
    if (ans > tmp) ans = tmp;
  }
  cout << ans << endl;
}


int main(){
  cin >> N >> X;
  for(li i = 0; i < N; i++){
    cin >> A[i];
    B.push_back(A[i]);
  }
  solve();
  return 0;
}
