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;
int main() {
  long long n, x;
  scanf("%lld %lld", &n, &x);
  vector<pair<long long, long long>> a(n);
  for (int i = 0; i < n; i++) {
    long long t;
    cin >> t;
    a[i] = make_pair(t, i);
  }
  sort(a.begin(), a.end());
  int total = 0;
  long long ans = 0;
  vector<bool> check(n, false);
  vector<int> h, e;
  while (total != n) {
    long long tsum = 0;
    for (int i = 0; i < n; ++i) {
      if (check[a[i].second] == false) {
        tsum += a[i].first;
        if (tsum > x && h.size() != 0) {
          tsum -= a[i].first;
        }
        h.push_back(i);
      }
    }
    ans += tsum;
    e.clear();
    for (int i = 0; i < n; i++) check[i] = false;
    for (auto itr : h) {
      if (itr == n - 1) {
        check[0] = true;
        e.push_back(0);
      } else {
        check[itr + 1] = true;
        e.push_back(itr + 1);
      }
    }
    h.clear();
    total = e.size();
    if (total != h.size()) {
      ans += x;
    }
  }
  cout << ans << endl;
}
