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 qp(int a, long long b) {
  int ans = 1;
  do {
    if (b & 1) ans = 1ll * ans * a;
    a = 1ll * a * a;
  } while (b >>= 1);
  return ans;
}
long long qp(int a, long long b, int mo) {
  int ans = 1;
  do {
    if (b & 1) ans = 1ll * ans * a % mo;
    a = 1ll * a * a % mo;
  } while (b >>= 1);
  return ans;
}
long long gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
long long INF = 1e14;
long long cir(vector<long long> a, int m) {
  long long n = a.size();
  if (m >= 0)
    return a[m];
  else
    return a[n - (-1 * m % n)];
}
int main(void) {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  vector<vector<long long>> b(n, vector<long long>(n, 0));
  long long ans = 1e16;
  for (int k = 0; k <= n - 1; k++) {
    int res = 0;
    for (int i = 0; i < n; i++) {
      if (k == 0)
        b[k][i] = a[i];
      else {
        b[k][i] = min(b[k - 1][i], cir(a, i - k));
      }
    }
    for (int i = 0; i < n; i++) {
      res += b[k][i];
    }
    ans = min(ans, res + k * x);
  }
  cout << ans << endl;
  return 0;
}
