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;
int main() {
  int64_t N, X;
  cin >> N >> X;
  vector<int64_t> generateCosts(N);
  vector<int64_t> moves(N);
  vector<int64_t> origins(N);
  vector<int64_t> leastCosts(N);
  for (int i = 0; i < N; i++) {
    cin >> generateCosts[i];
    leastCosts[i] = generateCosts[i];
    moves[i] = 0;
    origins[i] = i;
  }
  for (int origin = 0; origin < N; origin++) {
    for (int move = 0; move < N; move++) {
      int target = origin + move;
      if (target >= N) target -= N;
      if (leastCosts[target] > generateCosts[origin] + X * move) {
        leastCosts[target] = generateCosts[origin] + X * move;
        moves[target] = move;
        origins[target] = origin;
      }
    }
  }
  int64_t maxMove = *max_element(moves.begin(), moves.end());
  int64_t result = maxMove * X;
  cerr << maxMove << endl;
  for (int v = 0; v < N; v++) {
    result += generateCosts[origins[v]];
  }
  cout << result << endl;
  return 0;
}
