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 ll = long long;
using vi = vector<int>;
void ins() {}
template <class T, class... Rest>
void ins(T& v, Rest&... rest) {
  cin >> v;
  ins(rest...);
}
int main() {
  int N;
  ll x;
  ins(N, x);
  vector<ll> A(N);
  for (int i = 0, _i = (N); i < _i; ++i) cin >> A[i];
  int count = 0;
  ll ans = 0;
  for (int i = 0, _i = (N); i < _i; ++i) {
    ll cost = A[i];
    int idx = i;
    for (int j = 0, _i = (N); j < _i; ++j) {
      ll C = A[j] + x * ((i - j + N) % N);
      if (cost > C) {
        cost = A[j] + x * ((i - j + N) % N);
        idx = j;
      } else if (cost == C && (i - j + N) % N > (i - idx + N) % N) {
        cost = A[j] + x * ((i - j + N) % N);
        idx = j;
      }
    }
    ans += A[idx];
    count = max(count, (i - idx + N) % N);
  }
  cout << ans + 1LL * count * x << endl;
  return 0;
}
