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;
const long long MOD = 1e9 + 7;
inline long long ni() {
  long long x;
  cin >> x;
  return x;
}
template <class T>
inline T nt() {
  T x;
  cin >> x;
  return x;
}
inline void print() {}
template <typename T, typename... TAIL>
inline void print(const T &t, TAIL... tail) {
  cout << t;
  print(tail...);
}
inline void input() { return; }
template <typename T, typename... TAIL>
inline void input(T &t, TAIL &...tail) {
  cin >> t;
  input(tail...);
}
const long long N = 2 * 1e5 + 1;
const long long inf = 1e18 + 7;
const double PI = 3.14159265359;
long long cost[N];
long long best[N];
signed main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  long long n, x;
  cin >> n >> x;
  vector<long long> v;
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    v.push_back(ni());
    res += v[i];
    cost[i] = v[i];
  }
  for (long long j = 0; j < n; j++) {
    for (long long i = 0; i < n; i++) {
      if (cost[i] > j * x + v[(i - j + n) % n]) {
        best[i] = j;
        cost[i] = j * x + v[(i - j + n) % n];
      }
    }
    long long ans = 0;
    long long mx = 0;
    for (long long i = 0; i < n; i++) {
      mx = max(mx, best[i]);
    }
    for (long long i = 0; i < n; i++) {
      ans += cost[i] - best[i] * x;
    }
    ans += j * x;
    res = min(res, ans);
  }
  cout << res << '\n';
  return 0;
}
