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;
using ll = long long;
using ii = pair<int, int>;
void solve();
clock_t start_time;
double get_current_time() {
  return (double)(clock() - start_time) / CLOCKS_PER_SEC;
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
}
const int iinf = (int)1e9 + 123123;
const ll linf = (ll)1e17 + 123123;
const int limit = 10050;
const int k = 15;
int st[limit][k];
int get_min(int l, int r) {
  int j = log2(r - l + 1);
  return min(st[l][j], st[r - (1 << j) + 1][j]);
}
void solve() {
  int n, x;
  cin >> n >> x;
  vector<int> a(n + n);
  for (int i = 0; i < n + n; i++) {
    if (i < n) {
      cin >> a[i];
    } else {
      a[i] = a[i - n];
    }
  }
  for (int i = 0; i < n + n; i++) {
    st[i][0] = a[i];
  }
  for (int j = 1; j < k; j++) {
    for (int i = 0; i + (1 << j) <= n + n; i++) {
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
    }
  }
  ll ans = linf;
  for (int i = 0; i < n; i++) {
    ll may = x * i;
    for (int j = n - 1; j < n + n - 1; j++) {
      may += get_min(j - i, j);
    }
    ans = min(ans, may);
  }
  cout << ans << endl;
}
