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;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int M = 1000000007;
template <class T>
void debug(T a) {
  for (auto i : a) cout << i << endl;
}
int main() {
  int n, x;
  cin >> n >> x;
  vector<int> a, X;
  for (int i = 0; i < (int)n; i++) {
    int k;
    cin >> k;
    ;
    a.push_back(k);
    X.push_back(0);
  }
  long long ans = 0;
  int p = 0;
  auto prev = [n](int i) -> int { return (i == 0 ? n - 1 : i - 1); };
  auto next = [n](int i) -> int { return (i == n - 1 ? 0 : i + 1); };
  for (int j = 0; j < (int)2; j++)
    for (int i = 0; i < (int)n; i++) {
      int pr = prev(i);
      if (a[i] > a[pr] + x) {
        a[i] = a[pr] + x;
        X[i] = X[pr] + 1;
        p = max(p, X[i]);
      }
    }
  ans += p * x;
  bool ok = false;
  for (int i = 0; i < (int)n; i++) {
    if (ok) break;
    if (X[i] == 0) {
      int c = 1, j = i;
      while (1) {
        if (X[next(j)] == 0) break;
        if (next(j) == 0) ok = true;
        j = next(j);
        c++;
      }
      ans += c * a[i];
      i = j;
    }
  }
  cout << ans << endl;
}
