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>
const double EPS = 1e-9;
const int N = 2000 + 10;
using namespace std;
void dbg() { cout << "\n"; }
template <typename T, typename... A>
void dbg(T a, A... x) {
  cout << a << ' ';
  dbg(x...);
}
int st[2 * N][25], n;
long long x, a[N], ans;
void init() {
  for (int i = 1; i <= 2 * n; i++) st[i][0] = a[i];
  for (int j = 1; (1 << j) <= 2 * n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= 2 * n; i++)
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  }
}
int query(int l, int r) {
  int k = (int)(log((double)(r - l + 1)) / log(2.0));
  return min(st[l][k], st[r - (1 << k) + 1][k]);
}
int main() {
  scanf("%d%lld", &n, &x);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = n + 1; i <= 2 * n; i++) a[i] = a[i - n];
  init();
  for (int i = 1; i <= n; i++) ans += a[i];
  for (int k = 1; k <= n; k++) {
    long long tmp = (long long)k * x;
    for (int i = 1; i <= n; i++) {
      int x = i + n - k, y = i + n;
      tmp += query(x, y);
    }
    ans = min(ans, tmp);
  }
  printf("%lld\n", ans);
  return 0;
}
