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;
const double PI = 3.14159265358979323846;
const double EPS = 1e-9;
const long long INF = 10e17;
double equal(double a, double b) { return fabs(a - b) < DBL_EPSILON; }
std::istream &operator>>(std::istream &in, set<string> &o) {
  string a;
  in >> a;
  o.insert(a);
  return in;
}
std::istream &operator>>(std::istream &in, queue<int> &o) {
  long long a;
  in >> a;
  o.push(a);
  return in;
}
bool contain(set<int> &s, int a) { return s.find(a) != s.end(); }
long long min_r(vector<long long> &slimes, int i, int k, int n) {
  long long ans = INF;
  for (int j = 0; j < (k + 1); ++j) {
    long long mi = slimes[(i + n - j) % n];
    ans = min(ans, mi);
  }
  return ans;
}
long long check(int k, int n, vector<long long> &slimes) {
  long long ans = 0;
  for (int i = 0; i < (n); ++i) {
    long long mi = min_r(slimes, i, k, n);
    ans += mi;
  }
  return ans;
}
int main() {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> slimes(n);
  for (int i = 0; i < (n); ++i) cin >> slimes[i];
  long long ans = INF;
  for (int k = 0; k < (n); ++k) {
    long long now = check(k, n, slimes);
    ans = min(ans, now + (x * k));
  }
  cout << ans << endl;
}
