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;
const int MOD = 1000000007;
const int N = 2005;
long long int dp[N][N], a[N];
long long int mina(long long int x, long long int y) {
  if (x < y) return x;
  return y;
}
int main() {
  long long int n, x;
  cin >> n >> x;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    dp[i][1] = a[i];
  }
  for (int j = 2; j <= n; j++) {
    dp[1][1] = mina(dp[1][1], (long long int)(n + 1 - j) * (long long int)x +
                                  (long long int)a[j]);
  }
  for (int i = 2; i <= n; i++) dp[i][1] = mina(dp[i][1], dp[i - 1][1] + x);
  dp[1][1] = min(dp[1][1], dp[n][1] + x);
  for (int len = 2; len <= n; len++) {
    for (int i = 1; i <= n; i++) {
      int temp = (i + len - 1) % n;
      if (temp == 0) temp = n;
      if (i != 1)
        dp[i][len] = mina(dp[i - 1][len] + x, dp[i][len - 1] + dp[temp][1]);
      else
        dp[i][len] = dp[i][len - 1] + dp[i + len - 1][1];
    }
    dp[1][len] = mina(dp[1][len], dp[n][len] + x);
  }
  long long int ans = 1000000000000000000LL;
  for (int i = 1; i <= n; i++) {
    ans = mina(ans, dp[i][n]);
  }
  cout << ans << endl;
  return 0;
}
