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 int MOD = 1000000007;
const int INF = INT_MAX / 3 - 1;
const double EPS = 1e-9;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int dy8[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dat[2 * 2000];
int n;
void segment_tree(int n_, int x) {
  n = 1;
  while (n < n_) n *= 2;
  for (int i = 0; i < 2 * n - 1; i++) dat[i] = x;
}
int F(int p, int q) { return min(p, q); }
void update(int k, int a) {
  k += n - 1;
  dat[k] = a;
  while (k > 0) {
    k = (k - 1) / 2;
    dat[k] = F(dat[k * 2 + 2], dat[k * 2 + 1]);
  }
}
int query(int a, int b, int k, int l, int r) {
  if (r <= a || b <= l) return INF;
  if (a <= l && r <= b)
    return dat[k];
  else {
    int v1 = query(a, b, k * 2 + 1, l, (l + r) / 2);
    int v2 = query(a, b, k * 2 + 2, (l + r) / 2, r);
    return F(v1, v2);
  }
}
int main() {
  int n;
  long long int x;
  cin >> n >> x;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  segment_tree(n, 1e9 + 1);
  for (int i = 0; i < n; i++) update(i, a[i]);
  long long int res = 1e15;
  for (int i = 0; i < n; i++) {
    long long int sum = 0;
    for (int j = 0; j < n; j++)
      if (j - i < 0)
        sum += min(query(0, j + 1, 0, 0, n), query(n + j - i, n, 0, 0, n));
      else
        sum += query(j - i, j + 1, 0, 0, n);
    if (res > sum + x * i) res = sum + x * i;
  }
  cout << res << endl;
  return 0;
}
