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 N;
long long X;
vector<long long> A;
long long sub(long long x, long long y) {
  if (x < y) x += N;
  return x - y;
}
signed main() {
  cin >> N >> X;
  for (long long i = 0; i < N; ++i) {
    long long z;
    cin >> z;
    A.push_back(z);
  }
  long long ans = LLONG_MAX;
  vector<vector<long long> > b(N);
  for (long long i = 0; i < N; ++i) {
    b[i].resize(N);
    fill(b[i].begin(), b[i].end(), 0);
    b[i][0] = A[i];
  }
  for (long long k = 0; k < N - 1; ++k) {
    long long sum_b = 0;
    for (long long i = 0; i < N; ++i) {
      b[i][k] = min(b[i][k - 1], A[sub(i, k)]);
      sum_b += b[i][k];
    }
    ans = min(ans, k * X + sum_b);
  }
  cout << ans << endl;
  return 0;
}
