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;
long long dp[2002][2002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long N;
  cin >> N;
  long long x;
  cin >> x;
  vector<long long> a(N, 0LL);
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
  }
  long long res = 1LL << 60;
  ;
  for (int i = 0; i < (N); i++) {
    for (int j = 0; j < (N + 1); j++)
      for (int k = 0; k < (N + 1); k++) dp[j][k] = 1LL << 60;
    ;
    dp[0][0] = a[i % N];
    for (int j = 1; j <= N; ++j)
      dp[j][0] = a[(i + j) % N] + (j != 0 ? dp[j - 1][0] : 0);
    for (int k = 1; k <= N; ++k) dp[0][k] = x + a[i % N] + dp[0][k - 1];
    for (int j = 1; j <= N; ++j) {
      for (int k = 1; k <= N; ++k) {
        dp[j][k] = min(dp[j][k], dp[j - 1][k] + a[(i + j) % N]);
        dp[j][k] = min(dp[j][k], dp[j][k - 1] + x + a[i % N]);
      }
    }
    for (int j = 0; j <= N; ++j) {
      res = min(res, dp[N - 1][j]);
      res = min(res, dp[j][N - 1]);
    }
  }
  cout << res << endl;
  return 0;
}
