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>
#pragma GCC optimize("Ofast")
using namespace std;
const int64_t INF = (1LL << 60);
const long double inf = pow(10, 10);
const long double pi = acosl((long double)1);
const int64_t mod = 1000000007;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
const int ddx[8] = {1, 0, -1, -1, -1, 0, 1, 1};
const int ddy[8] = {1, 1, 1, 0, -1, -1, -1, 0};
void solve() {
  int64_t n, x;
  cin >> n >> x;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < (int64_t)(n); i++) cin >> a[i];
  vector<vector<int64_t>> dp(n, vector<int64_t>(n + 2, INF));
  for (int64_t i = 0; i < (int64_t)(n); i++) dp[i][0] = a[i];
  for (int64_t i = 0; i < (int64_t)(n); i++) {
    for (int64_t j = 0; j < (int64_t)(n); j++) {
      dp[i][j + 1] = min(a[(i - (j) + 3 * n) % n], dp[i][j]);
    }
  }
  int64_t ans = INF;
  for (int64_t j = 0; j < (int64_t)(n + 1); j++) {
    int64_t num = j * x;
    for (int64_t i = 0; i < (int64_t)(n); i++) {
      num += dp[i][j];
    }
    ans = min(ans, num);
  }
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  cout << setprecision(30);
  solve();
}
