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;
struct State {
  long long sec;
  long long spells;
  long long origin;
  bool operator<(const State& r) const {
    return sec != r.sec ? sec < r.sec : spells < r.spells;
  }
};
State memo[2][2001];
int main() {
  long long N, x;
  cin >> N >> x;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2001; j++) {
      memo[i][j] = State{LLONG_MAX, LLONG_MAX, -1};
    }
  }
  vector<long long> A(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  int curr = 0;
  for (long long k = 0; k < N; k++) {
    for (long long i = 0; i < N; i++) {
      if (k > 0) {
        memo[curr ^ 1][i] = memo[curr][i];
      }
      memo[curr ^ 1][i] =
          min(memo[curr ^ 1][i],
              State{A.at((i - k + N) % N) + k * x, k, (i - k + N) % N});
    }
    curr ^= 1;
  }
  long long km = 0;
  for (int i = 0; i < N; i++) {
    km = max(km, memo[curr][i].spells);
  }
  long long sum = km * x;
  for (int i = 0; i < N; i++) {
    sum += A.at(memo[curr][i].origin);
  }
  cout << sum << endl;
  return 0;
}
