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 long long N = 2007;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, x;
  cin >> n >> x;
  long long a[n + 1];
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  long long dp[n + 1][n + 1][n + 1];
  for (long long i = 0; i <= n; ++i) {
    for (long long j = 0; j <= n; ++j) {
      for (long long k = 0; k <= n; ++k) {
        dp[i][j][k] = 1e18;
      }
    }
  }
  dp[0][0][0] = 0;
  for (long long i = 0; i < n; ++i) {
    for (long long cnt = 0; cnt <= n; ++cnt) {
      for (long long mx = 0; mx <= n; ++mx) {
        for (long long j = 0; j <= n; ++j) {
          long long new_i = i + 1;
          long long new_cnt = cnt + j;
          long long new_mx = max(mx, j);
          if (new_i <= n && new_cnt <= n && new_mx <= n) {
            dp[new_i][new_cnt][new_mx] =
                min(dp[new_i][new_cnt][new_mx], dp[i][cnt][mx] + a[new_i] * j);
          }
        }
      }
    }
  }
  long long ans = 1e18;
  for (long long i = 1; i <= n; ++i) {
    long long val = dp[n][n][i] + (i - 1) * x;
    ans = min(ans, val);
  }
  cout << ans;
}
