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 int const MOD = 1000000007;
struct mypair {
  long long int a;
  long long int b;
  mypair() {}
  mypair(long long int i, long long int j) {
    a = i;
    b = j;
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long int n, x;
  cin >> n >> x;
  vector<mypair> a(n);
  for (int i = 0; i < n; i++) {
    long long int k;
    cin >> k;
    a[i] = mypair(k, 0);
  }
  for (int i = 0; i < n; i++) {
    if (a[(i + 1) % n].a > a[i].a + x) {
      a[(i + 1) % n].a = a[i].a + x;
      a[(i + 1) % n].b = a[i].b + 1;
    }
  }
  long long int sum = 0;
  long long int maxk = 0;
  long long int sumb = 0;
  for (int i = 0; i < n; i++) {
    sum += a[i].a;
    sumb += a[i].b;
    if (maxk < a[i].b) {
      maxk = a[i].b;
    }
  }
  cout << sum - sumb * x + maxk * x << endl;
  return 0;
}
