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 MAX_N = 5000;
const long long INF = (1LL << 31) - 1;
long long dat[2 * MAX_N], n;
void init(long long a) {
  n = 1LL;
  while (n < a) {
    n *= 2;
  }
  for (long long i = 0; i < 2 * MAX_N; 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[2 * k + 1], dat[2 * k + 2]);
  }
}
long long find(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 v1 = find(a, b, 2 * k + 1, l, (l + r) / 2);
    long long v2 = find(a, b, 2 * k + 2, (l + r) / 2, r);
    return min(v1, v2);
  }
}
long long N, x;
signed main(void) {
  cin >> N >> x;
  init(N + 1);
  for (long long i = 0; i < N; i++) {
    long long a;
    cin >> a;
    update(i, a);
  }
  long long ans = 1e9;
  for (long long k = 0; k < N; k++) {
    long long res = 0;
    for (long long i = 0; i < N; i++) {
      if (i - k >= 0) {
        res += find(i - k, i + 1, 0, 0, n);
      } else {
        long long r1 = find(0, i + 1, 0, 0, n);
        long long r2 = find(i - k + N, N, 0, 0, n);
        res += min(r1, r2);
      }
    }
    ans = min(ans, res + k * x);
  }
  cout << ans << endl;
  return 0;
}
