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;
inline bool check(long long x, long long y, long long xMax, long long yMax) {
  return x >= 0 && y >= 0 && xMax > x && yMax > y;
}
inline int toint(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class tuple<long long, long long, long long> >
inline string tostring(tuple<long long, long long, long long> x) {
  ostringstream sout;
  sout << x;
  return sout.str();
} template <class tuple<long long, long long, long long> >
inline tuple<long long, long long, long long> sqr(
    tuple<long long, long long, long long> x) {
  return x * x;
} template <class tuple<long long, long long, long long> >
inline tuple<long long, long long, long long> mypow(
    tuple<long long, long long, long long> x, long long n) {
  tuple<long long, long long, long long> res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
} inline long long gcd(long long a, long long b) {
  return b ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const long long dx[8] = {1, 0, -1, 0, 1, 1, -1, -1},
                dy[8] = {0, -1, 0, 1, -1, 1, -1, 1};
const long long mod = 1e9 + 7;
const long long INF = 1e17 + 9;
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (long long(i) = (0); i < (n); (i)++) {
    ans += a[i];
  }
  for (long long i = 1; i <= n; i++) {
    long long res = i * x;
    for (long long j = 0; j < n; j++) {
      long long k = (j - 1);
      if (k == -1) k = n - 1;
      a[k] = min(a[k], a[j]);
      res += a[k];
    }
    ans = min(ans, res);
  }
  cout << ans << endl;
  return 0;
}
