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<long long> a(n);
  for (auto& youso_ : (a)) cin >> youso_;
  long long minnum = LINF;
  int pos = -1;
  vector<long long> numvec(n * 2, INF);
  vector<long long> difvec(n * 2, INF);
  a.insert(a.end(), (a).begin(), (a).end());
  for (int i = 0; i < (n * 2); ++i) {
    if (minnum + (i - pos) * x >= a[i]) {
      minnum = a[i];
      pos = i;
    }
    numvec[i] = min(minnum, numvec[i]);
    difvec[i] = min((long long)i - pos, difvec[i]);
  }
  for (int i = 0; i < (n); ++i) {
    if (numvec[i] + difvec[i] * x > numvec[i + n] + difvec[i + n] * x) {
      numvec[i] = numvec[i + n];
      difvec[i] = difvec[i + n];
    } else if (numvec[i] + difvec[i] * x == numvec[i + n] + difvec[i + n] * x)
      if (difvec[i] > difvec[i + n]) {
        numvec[i] = numvec[i + n];
        difvec[i] = difvec[i + n];
      }
  }
  numvec.resize(n);
  difvec.resize(n);
  cout << (accumulate((numvec).begin(), (numvec).end(), 0LL) +
           *max_element((difvec).begin(), (difvec).end()) * x)
       << endl;
  return 0;
}
