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;
const long long INF = 1LL << 62;
int main() {
  int N;
  cin >> N;
  long long x;
  cin >> x;
  vector<long long> a(N);
  vector<bool> used(N, false);
  for (int i = 0; i < (int)(N); i++) cin >> a[i];
  long long ans = 0;
  int shift = 0;
  for (int k = 0;; k++) {
    vector<pair<long long, int> > work;
    for (int i = 0; i < N; i++) {
      if (used[i] == false) work.push_back(make_pair(a[i], i));
    }
    if (work.empty()) break;
    sort(work.begin(), work.end());
    int start = work[0].second;
    int cnt = 0;
    used[start] = true;
    ans += a[start];
    for (int i = 1; i < N; i++) {
      if (a[start] + x * (long long)i < a[(start + i) % N]) {
        used[(start + i) % N] = true;
        cnt = i;
      } else
        break;
    }
    ans += a[start] * (long long)cnt;
    shift = max(shift, cnt);
  }
  ans += (long long)shift * x;
  cout << ans << endl;
  return 0;
}
