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;
long long n, x, a[2011], ans, kik;
bool used[2011];
pair<long long, bool> get(long long i) {
  long long sum = 0;
  bool check = 0;
  if (i == n) {
    if (a[i] <= a[1] + x) {
      sum = a[i];
    } else {
      sum = (a[1] + x);
      check = 1;
    }
  } else {
    if (a[i] <= a[i + 1] + x) {
      sum = a[i];
    } else {
      sum = (a[i + 1] + x);
      check = 1;
    }
  }
  return make_pair(sum, check);
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> x;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  long long ct = 0;
  for (int Q = 1; Q <= n; Q++) {
    bool kek = 0;
    long long tp = 0;
    long long mn = (long long)1e15;
    for (int I = 1; I <= n; ++I) {
      if (used[I]) continue;
      int i = (I + kik) % n;
      if (i == 0) {
        i = n;
      }
      pair<long long, bool> sum = get(i);
      if (sum.first < mn) {
        mn = sum.first;
        tp = I;
        kek = sum.second;
      }
    }
    used[tp] = 1;
    ans += mn;
    if (kek) {
      kik++;
    }
  }
  cout << ans;
  return 0;
}
