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;
using namespace std;
long long sum = 0;
long long x;
template <class T>
inline bool chmin(T& a, T b) {
  if (a >= b) {
    a = b - x;
    sum += x + a;
    return true;
  }
  sum += a;
  return false;
}
int main() {
  long long n;
  cin >> n >> x;
  vector<long long> a(n);
  set<pair<long long, long long> > se;
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
    se.insert(make_pair(a[i], i));
  }
  pair<long long, long long> p = *se.begin();
  sum = p.first;
  for (long long i = 0; i < (n - 1); i++) {
    if (p.second + i + 1 < n) {
      chmin(a[p.second + i + 1], a[p.second + i] + x);
    } else if (p.second + i + 1 == n) {
      chmin(a[p.second + i + 1 - n], a[p.second + i] + x);
    } else {
      chmin(a[p.second + i + 1 - n], a[p.second + i - n] + x);
    }
  }
  cout << sum << endl;
}
