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;
int N;
long long x, A[2000], B[2000];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(9);
  cin >> N >> x;
  long long m = 2147483647;
  int p = -1;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    if (A[i] < m) {
      m = A[i];
      p = i;
    }
  }
  for (int i = 0; i < N; i++) {
    B[(i - p + N) % N] = A[i];
  }
  long long ans = B[0];
  long long c = B[0];
  long long bef = B[0];
  long long maxX = 0;
  long long nowX = 0;
  for (int i = 1; i < N; i++) {
    if (bef + x < B[i]) {
      nowX++;
      maxX = max(maxX, nowX);
      ans += c;
      bef += x;
    } else {
      nowX = 0;
      ans += B[i];
      c = B[i];
      bef = B[i];
    }
  }
  cout << ans + x * maxX << endl;
  return 0;
}
