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;
mt19937 rnd(time(0) + 228 + 'k' + 'e' + 'k' + 'e' + 'r' + 'o' + 'f' + 'e' +
            'y');
const long double EPS = 1e-10;
bool checkprime(int x) {
  for (int i = 2; i * i <= x; ++i)
    if (x % i == 0) return 0;
  return 1;
}
const long double PI = acos(-1);
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
int mod = MOD9;
const int inf = 1e9;
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  srand(time(NULL));
  long long n, x;
  cin >> n >> x;
  long long ans = 1e18;
  vector<vector<long long>> mn(n, vector<long long>(n));
  vector<long long> a(n);
  for (long long i = 0; i < a.size(); ++i) cin >> a[i];
  ;
  for (long long i = 0; i < n; ++i) {
    long long mna = a[i];
    for (long long j = i; j < n; ++j) {
      mn[i][j] = mna;
      mna = min(mna, a[j]);
    }
  }
  for (long long am = 0; am <= n; ++am) {
    long long now = 1ll * am * x;
    for (long long i = 0; i < n; ++i) {
      if (i < am) {
        long long tx = mn[0][i];
        tx = min(tx, mn[(n - am + i)][n - 1]);
        now += tx;
      } else {
        long long tx = mn[i - am][i];
        now += tx;
      }
    }
    ans = min(ans, now);
  }
  cout << ans;
}
