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 ll = long long;
struct edge {
  ll to, cost;
};
std::vector<edge> v[100010], w[100010];
ll min(ll a, ll b) {
  if (a >= b)
    return b;
  else
    return a;
}
ll max(ll a, ll b) {
  if (a >= b)
    return a;
  else
    return b;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
  ll g = gcd(a, b);
  return a / g * b;
}
const ll Z = 1000000007;
const ll INF = 1 << 30;
const ll INF2 = 9000000000000000000LL;
bool fl = true;
bool f = true;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
std::string abc = "abcdefghijklmnopqrstuvwxyz";
int main() {
  ll n, x, a[10000], ans = INF2;
  std::cin >> n >> x;
  for (int i = 0; i < n; i++) {
    std::cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    ll cost = 0;
    for (int j = 0; j < n; j++) {
      ll k = j - i, mi = INF;
      if (k < 0) {
        for (int l = n + k; l < n; l++) {
          mi = min(a[l], mi);
        }
        k = 0;
      }
      for (; k <= j; k++) {
        mi = min(a[k], mi);
      }
      cost += mi;
    }
    ans = min(ans, cost + x * i);
  }
  std::cout << ans << std::endl;
}
