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 mod = 1000000007;
const long long inf = 1ll << 61;
const long double pi = 3.1415926535897932384626433;
struct edge {
  long long to, cost;
};
long long a[8005];
long long dat[280005], sdat[280005];
long long N;
void init(long long n_) {
  N = 1;
  while (N < n_) N *= 2;
  for (long long i = 0; i < 2 * N - 1; i++) {
    dat[i] = inf;
  }
}
void update(long long k, long long a) {
  k += N - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
  }
}
long long query(long long a, long long b, long long k, long long l,
                long long r) {
  if (r <= a || b <= l) return inf;
  if (a <= l && r <= b)
    return dat[k];
  else {
    long long vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
    long long vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return min(vl, vr);
  }
}
signed main() {
  long long n, x;
  cin >> n >> x;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) a[i + n] = a[i];
  n += n;
  long long ans = inf;
  init(n);
  for (long long i = 0; i < n; i++) update(i, a[i]);
  for (long long k = 0; k < n / 2; k++) {
    long long sum = 0;
    for (long long i = n / 2; i < n; i++) {
      long long s = i - k, t = i;
      sum += query(s, t + 1, 0, 0, n);
    }
    ans = min(ans, sum + k * x);
  }
  cout << ans << endl;
}
