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;
int n, x, k;
long long a[2048];
long long s(int i, int c, int d, int m) {
  if (i == n) return max(d, m) * x;
  long long b = a[(k + i) % n];
  long long r = c + s(i + 1, c, d + 1, m);
  long long r2 = b + s(i + 1, b, 0, max(d, m));
  return min(r, r2);
}
int main() {
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long m = 10000000001;
  for (int i = 0; i < n; i++)
    if (a[i] < m) {
      m = a[i];
      k = i;
    }
  stack<tuple<long long, long long> > s;
  s.push(make_tuple(n, 0));
  for (int i = n - 1; i >= 0; i--) {
    long long b = a[(k + i) % n];
    while (true) {
      long long c, d;
      tie(c, d) = s.top();
      if (c == n) break;
      s.pop();
      long long e = get<0>(s.top());
      long long f = (e - c) * d + (e - c - 1) * x;
      long long g = (e - c) * (b + x);
      if (g <= f) continue;
      s.push(make_tuple(c, d));
      break;
    }
    s.push(make_tuple(i, b));
  }
  long long z = 0, c, d, e, f;
  tie(c, d) = s.top();
  s.pop();
  while (true) {
    tie(e, f) = s.top();
    z += (e - c) * d + (e - c - 1) * x;
    if (e == n) break;
    c = e;
    d = f;
    s.pop();
  }
  cout << z << endl;
}
