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>
const int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const int dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using namespace std;
long long max(long long a, int b) { return max(a, long long(b)); }
long long max(int a, long long b) { return max(long long(a), b); }
long long min(long long a, int b) { return min(a, long long(b)); }
long long min(int a, long long b) { return min(long long(a), b); }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, x;
  cin >> n >> x;
  vector<long long> v(n);
  for (long long i = 0; i < long long(n); ++i) cin >> v[i];
  vector<vector<int> > q(n);
  for (long long i = 0; i < long long(n); ++i) {
    int p = -1;
    long long cmin = 1000000001ll * 1000000001ll;
    for (long long j = 0; j < long long(n); ++j) {
      if (cmin > v[(i + j) % n] + j * x) {
        cmin = v[(i + j) % n] + j * x;
        p = (i + j) % n;
      }
    }
    q[(i - p + n) % n].push_back(p);
  }
  long long ans = 0, cnt = 0;
  for (long long i = long long(n) - 1; i >= 0; --i) {
    if (cnt != 0) ans += x;
    for (long long j = 0; j < long long(q[i].size()); ++j) {
      cnt++;
      ans += v[q[i][j]];
    }
  }
  cout << ans << endl;
}
