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 int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
const long long MOD = (long long)1e9 + 7;
long long a[2004], b[2004], c[2004];
int main() {
  long long N, x;
  cin >> N >> x;
  for (int i = (0); i < (N); ++i) {
    cin >> a[i];
    b[i] = a[i];
    c[i] = i;
  }
  int cnt = 0;
  for (int i = (1); i < (3 * N); ++i) {
    if (b[i % N] > b[(i - 1) % N] + x) {
      b[i % N] = b[(i - 1) % N] + x;
      c[i % N] = c[(i - 1) % N];
    }
  }
  long long res = 0;
  long long maxi = 0;
  for (int i = (0); i < (N); ++i) {
    res += a[c[i]];
    maxi = max(maxi, ((N + i - c[i]) % N) * x);
  }
  cout << res + maxi << endl;
  return 0;
}
