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>
const int INF = 1e9;
const int MOD = 1e9 + 7;
const long long LINF = 1e18;
using namespace std;
void CINT() {}
template <class Head, class... Tail>
void CINT(Head&& head, Tail&&... tail) {
  cin >> head;
  CINT(move(tail)...);
}
int main() {
  int n, x;
  CINT(n, x);
  vector<int> a(n);
  for (auto& youso_ : (a)) cin >> youso_;
  pair<int, int> minnum = make_pair(INF, -1);
  vector<pair<int, int> > saitan(n * 2, make_pair(INF, -1));
  for (int i = 0; i < (n * 2); ++i) {
    if (minnum.first + (i - minnum.second) * x > a[i % n])
      minnum = make_pair(a[i % n], i);
    saitan[i] = make_pair(minnum.first, (i - minnum.second));
  }
  for (int i = 0; i < (n); ++i) {
    if (saitan[i].first + saitan[i].second * x >
        saitan[i + n].first + saitan[i + n].second * x)
      saitan[i] = saitan[i + n];
    else if (saitan[i].first + saitan[i].second * x ==
             saitan[i + n].first + saitan[i + n].second + x) {
      if (saitan[i].second > saitan[i + n].second) saitan[i] = saitan[i + n];
    }
  }
  int maxel = 0;
  long long ans = 0;
  for (int i = 0; i < (n); ++i) {
    maxel = max(maxel, saitan[i].second);
    ans += saitan[i].first;
  }
  cout << (ans + maxel * x) << endl;
  return 0;
}
