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;
long long N, x;
long long a[2001];
vector<pair<long long, long long> > vv;
long long calc(const vector<bool>& now) {
  long long rtn = 0;
  for (int i = 0; i < now.size(); ++i)
    if (now[i]) rtn += (a[(i + 1) % N] - a[i]);
  return rtn;
}
int main() {
  scanf("%lld %lld", &N, &x);
  for (int i = 0; i < N; ++i) {
    scanf("%lld", a + i);
    vv.emplace_back(a[i], i);
  }
  sort(vv.begin(), vv.end());
  vector<bool> now(N, 0);
  long long ans = 0, mycnt = 0;
  for (int i = 0; i < vv.size(); ++i)
    if (vv[i].first <= x) {
      now[i] = true;
      ans += vv[i].first;
      ++mycnt;
    }
  while (mycnt < N) {
    if (calc(now) > x) {
      ans += x;
      rotate(now.begin(), now.begin() + N - 1, now.end());
    } else {
      for (int i = 0; i < vv.size(); ++i)
        if (!now[vv[i].second]) {
          now[vv[i].second] = true;
          ans += vv[i].first;
          ++mycnt;
          break;
        }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
