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;
long long t[2001][2001];
long long N, x;
vector<long long> rest_transform[2001];
long long result = 0;
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; i++) {
    long long a;
    cin >> a;
    for (int j = 0; j < N; j++) {
      t[i][(i + j) % N] = a + j * x;
    }
  }
  long long interval_max = 0;
  for (int j = 0; j < N; j++) {
    long long min_sec = 1 << 30;
    long long i_min;
    for (int i = 0; i < N; i++) {
      if (t[i][j] < min_sec) {
        min_sec = t[i][j];
        i_min = i;
      }
    }
    rest_transform[(N + j - i_min) % N].push_back(t[i_min][i_min]);
    if ((N + j - i_min) % N > interval_max) {
      interval_max = (N + j - i_min) % N;
    }
  }
  for (int i = interval_max; i >= 0; i--) {
    for (auto it = rest_transform[i].begin(); it != rest_transform[i].end();
         it++) {
      result += *it;
    }
    result += x;
  }
  result -= x;
  cout << result << endl;
  return (0);
}
