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 maxn = 1e5 + 100;
const long long inf = 2e18, mod = 1e9 + 7;
int n, mlt[maxn];
long long x, a[maxn], ans = inf;
pair<long long, int> valind[maxn];
bool selection[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> a[i], valind[i] = make_pair(a[i], i);
  sort(valind, valind + n);
  for (int startCount = 0; startCount < n; startCount++) {
    memset(mlt, 0, sizeof(mlt));
    selection[valind[startCount].second] = true;
    int pos = valind[startCount].second, cnt = 0;
    for (int i = (pos + 1) % n; i != pos; i = (i + 1) % n) {
      cnt++;
      if (selection[i] == true) mlt[i] = cnt, cnt = 0;
    }
    cnt++;
    mlt[pos] = cnt;
    long long cur = *max_element(mlt, mlt + n) * x - x;
    for (int i = 0; i < n; i++)
      if (selection[i]) cur += a[i] * mlt[i];
    ans = min(ans, cur);
  }
  cout << ans << endl;
}
