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 b[2000][2000];
void solve() {
  int n, x;
  cin >> n >> x;
  int a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  pair<long long, int> ans[n];
  for (int i = 0; i < n; ++i) ans[i] = {a[i], 0};
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      b[i][j] = (long long)(i - j) * x + a[j];
      if (ans[i].first > b[i][j]) {
        ans[i] = {b[i][j], i - j};
      }
    }
  }
  for (int j = 1; j < n; ++j) {
    b[0][j] = (long long)(n - j) * x + a[j];
    if (ans[0].first > b[0][j]) {
      ans[0] = {b[0][j], n - j};
    }
  }
  long long c = 0;
  for (int i = 0; i < n; ++i) c += ans[i].first;
  int mmax = 0, sum = 0;
  for (int i = 0; i < n; ++i) {
    mmax = (mmax >= ans[i].second ? mmax : ans[i].second);
    sum += ans[i].second;
  }
  c -= (sum - mmax) * x;
  cout << c << endl;
}
int main() { solve(); }
