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;
double EPS = 1e-9;
int INFi = 1000000005;
long long INFll = 1000000000000000005ll;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
long long MOD = 1000000007;
const int MAX_N = 1 << 17;
long long n, dat[2 * MAX_N - 1];
void init(int n_) {
  n = 1;
  while (n < n_) n *= 2;
  for (int i = 0; i < 2 * n - 1; i++) dat[i] = INFi;
}
void update(int 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(int a, int b, int k, int l, int r) {
  if (r <= a || b <= l) return INFi;
  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);
  }
}
int N;
long long x;
long long a[MAX_N];
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; i++) cin >> a[i];
  init(2 * N);
  for (int i = 0; i < N; i++) {
    update(i, a[i]);
    update(i + N, a[i]);
  }
  long long ans = INFll;
  for (int i = 0; i < N; i++) {
    long long s = i * x;
    for (int j = 0; j < N; j++) {
      s += query(N + j - i, N + j + 1, 0, 0, 2 * N);
    }
    ans = min(ans, s);
  }
  cout << ans << endl;
  return 0;
}
