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;
static const long long MAX = 2001;
long long dp[MAX][MAX] = {};
int main() {
  long long n, x;
  cin >> n >> x;
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    long long p;
    cin >> p;
    a.push_back(p);
  }
  for (long long j = 0; j < n; j++) {
    dp[1][j] = a[j];
    if (j) {
      dp[2][j] = min(a[j], a[j - 1]);
    } else {
      dp[2][j] = min(a[j], a[n - 1]);
    }
  }
  for (long long i = 2; i <= n; i++) {
    for (long long j = 0; j < n; j++) {
      if (j - 1 >= 0 and j - i >= 0) {
        dp[i + 1][j] = min(dp[i - 1][j - 1], min(dp[1][j], dp[1][j - i]));
      } else if (j - i >= 0) {
        dp[i + 1][j] = min(dp[i - 1][j - 1 + n], min(dp[1][j], dp[1][j - i]));
      } else if (j - 1 >= 0) {
        dp[i + 1][j] = min(dp[i - 1][j - 1], min(dp[1][j], dp[1][j - i + n]));
      } else {
        dp[i + 1][j] =
            min(dp[i - 1][j - 1 + n], min(dp[1][j], dp[1][j - i + n]));
      }
    }
  }
  long long ans;
  for (long long num_magic = 0; num_magic < n; num_magic++) {
    long long pre = 0;
    for (long long j = 0; j < n; j++) {
      pre += dp[num_magic + 1][j];
    }
    pre += num_magic * x;
    if (num_magic == 0) {
      ans = pre;
    } else {
      if (ans > pre) {
        ans = pre;
      }
    }
  }
  cout << ans << endl;
}
