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;
using cd = complex<double>;
const int N = 3e3 + 5;
const long long INF = 1e18;
const int mod = 998244353;
const double Pi = acos(-1);
void Fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int n;
long long a[6005], b[8005];
long long ans = 0, x = 0;
void Solve(int k) {
  deque<pair<int, int> > Q;
  for (int i = 1; i <= k; i++) {
    while (Q.empty() == false) {
      if (a[i] < Q.front().first) {
        Q.pop_front();
      } else {
        break;
      }
    }
    Q.push_back(make_pair(a[i], i));
  }
  b[1] = Q.front().first;
  for (int i = k + 1; i <= n + k - 1; i++) {
    while (Q.empty() == false) {
      if (Q.front().first > a[i]) {
        Q.pop_front();
      } else if (Q.front().second <= i - k) {
        Q.pop_front();
      } else {
        break;
      }
    }
    Q.push_back(make_pair(a[i], i));
    b[i - k + 1] = Q.front().first;
  }
}
signed main() {
  cin >> n >> x;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    ans += a[i];
  }
  for (int i = n + 1; i < 2 * n; i++) {
    a[i] = a[i - n];
  }
  for (int i = 0; i < n; i++) {
    Solve(i + 1);
    long long temp = x * i;
    for (int i = 1; i <= n; i++) {
      temp += b[i];
    }
    ans = min(ans, temp);
  }
  cout << ans;
}
