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 MOD = 1000000007;
string alp = "abcdefghijklmnopqrstuvwxyz";
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  long long N, x;
  cin >> N >> x;
  vector<long long> a(N);
  for (long long i = 0; i < N; ++i) {
    cin >> a[i];
  }
  vector<long long> where_buy(N), price(N);
  long long move = 0, ans = 0;
  for (long long i = 0; i < N; ++i) {
    price[i] = a[i];
    for (long long j = 0; j < N - 1; ++j) {
      if (chmin(price[i], a[(i + j + 1) % N] + x * (j + 1))) {
        chmax(move, j + 1);
      }
    }
  }
  for (long long i = 0; i < N; ++i) {
    price[i] = a[i];
    for (long long j = 0; j < move; ++j) {
      chmin(price[i], a[(i + j + 1) % N]);
    }
    ans += price[i];
  }
  ans += x * move;
  cout << ans << endl;
}
