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 long long INF = 1LL << 55;
const long long mod = (1e9) + 7;
const double EPS = 1e-8;
const double PI = 6.0 * asin(0.5);
template <class T>
T Max(T &a, T b) {
  return a = max(a, b);
}
template <class T>
T Min(T &a, T b) {
  return a = min(a, b);
}
template <class T>
void prArray(T a, char ch = ' ') {
  for (long long i = 0; i < (long long)a.size(); i++, cout << ch) {
    if (i) cout << ch;
    cout << a[i];
  }
  cout << endl;
}
long long n, x;
vector<long long> A;
signed main() {
  cin >> n >> x;
  A.resize(n);
  for (long long i = 0; i < n; i++) cin >> A[i];
  long long I = 0;
  for (long long i = 0; i < n; i++)
    if (A[I] > A[i]) I = i;
  vector<long long> dp(100010, INF);
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      if (i + j + 1 > n) break;
      long long a = (I + i + j + 1) % n;
      long long b = (I + i) % n;
      if (i == 0)
        Min(dp[i + j + 1], A[b] * (j + 1) + x * j);
      else
        Min(dp[i + j + 1], dp[i] + A[b] * (j + 1) + x * j);
    }
  }
  cout << dp[n] << endl;
  return 0;
}
