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;
template <typename X>
bool exist(vector<X> vec, X item) {
  return find(vec.begin(), vec.end(), item) != vec.end();
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long ans[2001][2001];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long N;
  cin >> N;
  long long x;
  cin >> x;
  vector<long long> a;
  for (long long i = (long long)0; i < (long long)N; i++) {
    long long inp;
    cin >> inp;
    a.push_back(inp);
  }
  for (long long i = (long long)0; i < (long long)N; i++)
    for (long long j = (long long)0; j < (long long)N; j++) {
      long long times = (j - i);
      if (times < 0) times += (N - 1);
      ans[i][j] = a[i] + times * x;
    }
  long long ret = 0;
  for (long long j = (long long)0; j < (long long)N; j++) {
    long long m = 100000000000000;
    for (long long i = (long long)0; i < (long long)N; i++) {
      m = min(m, ans[i][j]);
    }
    ret += m;
  }
  cout << ret << endl;
  return 0;
}
