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 <stdio.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <vector>
using namespace std;

//debug
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

int solve(int N, int x, vector<int> a) {
  int INT_MAX = 2147483647;
  int allMin = INT_MAX;
  for (int magic = 0; magic < N; magic++) {
    int sum = 0;
    for (int i = 0; i < N; i++) {
      int m = INT_MAX;
      int minCheck = i - magic;
      if (minCheck < 0) {
        minCheck += N;
      }

      do {
        if (minCheck == N) {
          minCheck = 0;
        }
        m = min(m, a[minCheck]);
        minCheck++;
      } while(minCheck != (i + 1));
      sum += m;
    }
    sum += x * magic;
    allMin = min(allMin, sum);
  }
  return allMin;
}

int main() {
  int N;
  cin >> N;
  int x;
  cin >> x;
  vector<int> list(N);
  for (int i = 0; i < N; i++) {
    cin >> list[i];
  }
  cout << solve(N, x, list)  << endl;
  /*
  int N = 3;
  int x = 10;
  vector<int> list;
  list.push_back(100);
  list.push_back(1);
  list.push_back(100);


  int N = 4;
  int x = 10;
  vector<int> list;
  list.push_back(1);
  list.push_back(2);
  list.push_back(3);
  list.push_back(4);
  cout << solve(N, x, list)  << endl;
  */
  return 0;
}
