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;
void mul_mod(long long& a, long long b) {
  a *= b;
  a %= (1000000007ll);
}
void add_mod(long long& a, long long b) {
  b += (1000000007ll);
  a += b;
  a %= (1000000007ll);
}
int n;
long long a[2222], memo[2222][2222];
bool done[2222][2222];
long long segmin(int l, int r) {
  if (done[l][r]) {
    return memo[l][r];
  }
  done[l][r] = true;
  if (l > r) {
    memo[l][r] = min(segmin(l, n - 1), segmin(0, r));
  } else if (l == r) {
    memo[l][r] = a[l];
  } else {
    memo[l][r] = min(a[l], segmin(l + 1, r));
  }
  return memo[l][r];
}
int main(void) {
  int i, j;
  long long x, tmp, ans = (1 << 30);
  fill(done[0], done[2222], false);
  cin >> n >> x;
  for (i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (i = 0; i < n; ++i) {
    tmp = i * x;
    for (j = 0; j < n; ++j) {
      tmp += segmin((j - i + n) % n, j);
    }
    ans = min(tmp, ans);
  }
  cout << ans << endl;
  return 0;
}
