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;
  } else {
    return 0;
  }
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  } else {
    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);
}
int main() {
  long long ans = 0;
  long long n, x;
  cin >> n >> x;
  long long a[2010];
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
  }
  long long mini[2010];
  for (int i = 0; i < (2010); i++) mini[i] = 100000000000000ll;
  long long sum = 0, most = 0;
  for (int i = 0; i < (n); i++) {
    long long v = 0;
    for (int j = 0; j < (n); j++) {
      if (mini[i] > ((i + n - j) % n) * x + a[j]) {
        mini[i] = ((i + n - j) % n) * x + a[j];
        v = ((i + n - j) % n) * x;
      }
    }
    sum += v;
    if (most < v) most = v;
  }
  for (int i = 0; i < (n); i++) {
    ans += mini[i];
  }
  ans -= sum;
  ans += most;
  cout << ans << endl;
}
