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 gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (a % b == 0)
    return b;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) {
  long long c = gcd(a, b);
  return ((a / c) * (b / c) * c);
}
const long long INF = 1e18 + 7;
const long long MOD = 1e9 + 7;
const long long N = 2003;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  long long mina = INF;
  for (long long i = (0); i < (n); ++i) {
    cin >> a[i];
    mina = min(mina, a[i]);
  }
  long long cost[n][n];
  for (long long i = (0); i < (n); ++i) cost[i][0] = a[i];
  for (long long i = (0); i < (n); ++i) {
    for (long long m = (1); m < (n); ++m) {
      long long l = i - m;
      if (l < 0) l += n;
      cost[i][m] = min(cost[i][m - 1], a[l]);
    }
  }
  vector<pair<long long, long long> > ac(n);
  for (long long i = (0); i < (n); ++i) {
    ac[i].second = i;
    ac[i].first = a[i];
  }
  (sort(ac.begin(), ac.end()));
  long long ans = mina * n + (n - 1) * x;
  for (long long m = (0); m < (n); ++m) {
    long long w = m * x;
    for (long long i = (0); i < (m); ++i) w += a[ac[i].second];
    for (long long i = (m); i < (n); ++i) {
      w += cost[ac[i].second][m];
    }
    ans = min(ans, w);
  }
  cout << (ans) << endl;
}
