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;
vector<int> A;
vector<vector<int> > dp;
int N, x;
int main() {
  cin >> N >> x;
  A.resize(N);
  dp.resize(N);
  for (int i = 0; i < (N); i++) {
    cin >> A[i];
    dp[i].resize(N);
  }
  for (int i = 0; i < (N); i++) {
    dp[i][0] = A[i];
    for (int j = (1); j < (N); j++) {
      int jj = i - j;
      if (jj < 0) {
        jj = N + jj;
      }
      if (dp[i][j - 1] > A[jj])
        dp[i][j] = A[jj];
      else
        dp[i][j] = dp[i][j - 1];
    }
  }
  long long ans = LLONG_MAX / 3;
  for (int i = 0; i < (N); i++) {
    long long ans0 = x * i;
    for (int j = 0; j < (N); j++) {
      ans0 += dp[j][i];
    }
    if (ans > ans0) ans = ans0;
  }
  cout << ans;
}
