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;
template <typename T>
inline void priv(vector<T> a) {
  for (int i = 0; i < a.size(); i++) {
    cerr << a[i] << ((i == a.size() - 1) ? "\n" : " ");
  }
}
long long gcd(long long a, long long b) {
  int c = max(a, b);
  int d = min(a, b);
  return c == 0 || d == 0 ? c : gcd(c % d, d);
}
long long lcm(long long a, long long b) {
  return a == 0 || b == 0 ? 0 : a * b / gcd(a, b);
}
long long fact(long long a) {
  long long b = 1;
  for (int i = 1; i <= a; i++) b *= i;
  return b;
}
long long power(long long x, long long n) {
  long long a = 1;
  for (int i = 0; i < n; i++) a *= x;
  return a;
}
int N, x;
long long a[2010];
int main() {
  cin >> N >> x;
  long long move = 0, ans = 0;
  for (int i = 0; i < N; i++) cin >> a[i];
  for (int i = 0; i < N; i++) {
    long long t1 = a[i], t2 = a[i];
    for (int j = 1; j <= N - 1; j++) {
      if (t1 > j * x + a[(i + (N - j)) % N]) {
        t1 = j * x + a[(i + (N - j)) % N];
        t2 = a[(i + (N - j)) % N];
      }
    }
    ans += t2;
    if (t2 != a[i]) move++;
  }
  cout << (ans + move * x) << endl;
  return 0;
}
