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 LONG2_MAX = 9223372036854775807;
long long min(long long a, long long b) { return a < b ? a : b; }
long long min2(int magic, int N, int index, vector<long long> a) {
  long long m = LONG2_MAX;
  int i = index - magic;
  if (i < 0) {
    i += N;
  }
  do {
    if (i == N) {
      i = 0;
    }
    m = min(m, a[i]);
    i++;
  } while (i != (index + 1));
  return m;
}
long long solve(int N, long long x, vector<long long> a) {
  long long allMin = LONG2_MAX;
  for (int magic = 0; magic < N; magic++) {
    long long sum = 0;
    long long memo[N];
    for (int i = 0; i < N; i++) {
      memo[i] = 0;
    }
    for (int i = 0; i < N; i++) {
      long long m = LONG2_MAX;
      ;
      if (magic == 0) {
        m = a[i];
      } else {
        long long prev = magic - 1;
        if (memo[prev] == 0) {
          m = min2(magic, N, prev, a);
          memo[prev] = m;
        } else {
          m = memo[prev];
        }
        m = min(m, a[i]);
      }
      sum += m;
    }
    sum += x * magic;
    allMin = min(allMin, sum);
  }
  return allMin;
}
int main() {
  int N;
  cin >> N;
  int x;
  cin >> x;
  vector<long long> list(N);
  for (int i = 0; i < N; i++) {
    cin >> list[i];
  }
  cout << solve(N, x, list) << endl;
  return 0;
}
