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 int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
const int dx_8[] = {1, 1, 0, -1, -1, -1, 0, 1},
          dy_8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T>
T GCD(T a, T b) {
  T c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
template <typename T>
T LCM(T a, T b) {
  T c = GCD(a, b);
  a /= c;
  return a * b;
}
template <typename T>
T nCr(T a, T b) {
  T i, r = 1;
  for (i = 1; i <= b; i++) {
    r *= (a + 1 - i);
    r /= i;
  }
  return r;
}
template <typename T>
T nHr(T a, T b) {
  return nCr(a + b - 1, b);
}
signed main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, x, a[2000];
  cin >> n >> x;
  for (long long i = 0; i < (n); ++i) cin >> a[i];
  long long ans = 0, change = 0;
  for (long long i = 0; i < (n); ++i) {
    long long MIN = 1LL * a[i];
    long long j = (i == 0 ? n - 1 : i - 1), cnt = 1;
    long long p = 0;
    while (j != i) {
      if (x * cnt + a[j] < MIN) {
        p = cnt;
        MIN = a[j];
      }
      j = (j == 0 ? n - 1 : j - 1);
      cnt++;
    }
    change = max(change, p);
    ans += MIN;
  }
  cout << ans + x * change << '\n';
  return 0;
}
