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;
constexpr long long mod = 1000000007;
long long n, x, a[2000], b[4000];
long long btr(long long l, long long r, long long i) {
  if (r - l == 1) return b[i] = a[l];
  b[i] = min(btr(l, (l + r) / 2, i * 2), btr((l + r) / 2, r, i * 2 + 1));
  return b[i];
}
long long rmq_(long long l, long long r, long long i, long long ql,
               long long qr) {
  if (ql <= l && qr >= r) return b[i];
  long long m = (l + r) / 2;
  long long v = 2000000000;
  if (ql < m) v = min(v, rmq_(l, m, i * 2, ql, qr));
  if (qr > m) v = min(v, rmq_(m, r, i * 2 + 1, ql, qr));
  return v;
}
long long rmq(long long l, long long r) {
  if (l >= r)
    return min(rmq_(0, n, 1, l, n), rmq_(0, n, 1, 0, r));
  else
    return rmq_(0, n, 1, l, r);
}
int main() {
  cin >> n >> x;
  for (long long i = 0; i < n; i++) cin >> a[i];
  btr(0, n, 1);
  long long r = 200000000000000;
  for (long long t = 0; t < n; t++) {
    long long c = t * x;
    for (long long s = 0; s < n; s++) {
      c += rmq((s - t + n) % n, s + 1);
    }
    r = min(r, c);
  }
  cout << r << endl;
}
