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 <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline T gcd(T a, T b) {
  T c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
const int INF = 1e9;
const double eps = 0.0000000001;
int main() {
  long long n, x;
  cin >> n >> x;
  long long a[10000];
  long long ans[10000];
  for (int(i) = 0; (i) < (n); (i)++) {
    cin >> a[i];
    ans[i] = a[i];
  }
  long long maxix = 0;
  for (int(i) = 0; (i) < (n); (i)++) {
    for (long long j = 1; j < n; j++) {
      if (ans[i] > a[(i + j) % n] + x * j) {
        ans[i] = a[(i + j) % n];
        chmax(maxix, j);
      }
    }
  }
  long long sum = 0;
  for (int(i) = 0; (i) < (n); (i)++) {
    sum += ans[i];
  }
  cout << sum + maxix * x << endl;
}
