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 UNDEF = -1;
const long long INF = 1e18;
template <typename T>
inline bool chkmax(T &aa, T bb) {
  return aa < bb ? aa = bb, true : false;
}
template <typename T>
inline bool chkmin(T &aa, T bb) {
  return aa > bb ? aa = bb, true : false;
}
const long long mn = 1e3 + 4;
long long a[mn];
int main() {
  ios_base::sync_with_stdio(false);
  long long n, rot;
  cin >> n >> rot;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long final = INF;
  for (long long k = 0; k < n; k++) {
    set<pair<long long, long long> > s;
    for (long long x = 0; x >= -k; x--) {
      long long y = (n + x) % n;
      s.insert(make_pair(a[y], y));
    }
    long long ans = 0;
    for (long long x = 0; x < n; x++) {
      long long cand = (s.begin())->first;
      ans += cand;
      long long y = (x - k + n) % n;
      s.erase(make_pair(a[y], y));
      s.insert(make_pair(a[x + 1], x + 1));
    }
    ans += k * rot;
    chkmin(final, ans);
  }
  cout << final << "\n";
}
