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 MOD = 1000000007;
const long long L_INF = 1LL << 60;
const long long INF = 2147483647;
const double PI = acos(-1);
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
void debug(T v) {
  for (long long i = 0; i < v.size(); ++i) cout << v[i] << " ";
  cout << endl;
}
const long long dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  auto b = a;
  long long res = L_INF;
  for (long long i = 0; i < n; ++i) {
    long long now = 0;
    for (long long j = 0; j < n; ++j) {
      long long idx = j - i;
      if (idx < 0) idx += n;
      b[j] = min(b[j], b[idx]);
      now += b[j];
    }
    now += x * i;
    res = min(res, now);
  }
  cout << res << endl;
}
