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 a[20001];
long long x;
long long now[20001];
vector<pair<long long, long long> > v;
int main() {
  cin >> N >> x;
  for (int i = 0; i < (N); i++) {
    cin >> a[i];
    v.push_back(pair<long long, long long>(a[i], i));
  }
  sort(v.begin(), v.end());
  long long ans = 1e17;
  for (int i = 0; i < (N); i++) {
    long long c = 0;
    vector<pair<long long, long long> > vv;
    for (int j = 0; j < (i + 1); j++) {
      vv.push_back(pair<long long, long long>(v[j].second, v[j].first));
      c += v[j].first;
    }
    sort(vv.begin(), vv.end());
    long long y = 0;
    for (int j = 0; j < (vv.size() - 1); j++) {
      y = max(vv[j + 1].first - vv[j].first - 1, y);
      c += ((vv[j + 1].first - vv[j].first - 1) * vv[j].second);
    }
    y = max(vv[0].first + N - vv[vv.size() - 1].first - 1, y);
    c += ((vv[0].first + N - vv[vv.size() - 1].first - 1) *
          vv[vv.size() - 1].second);
    ans = min(c + y * x, ans);
  }
  cout << ans << endl;
  return (0);
}
