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;
template <class T>
inline void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
template <class T>
inline void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T = long long>
T in() {
  T x;
  cin >> x;
  return (x);
}
struct Fast {
  Fast() {
    std::cin.tie(0);
    ios::sync_with_stdio(false);
  }
} fast;
template <typename T>
using PQ = priority_queue<T, vector<T>, greater<T>>;
long long dx[] = {1, -1, 0, 0};
long long dy[] = {0, 0, 1, -1};
long long solve() {
  long long N, x;
  cin >> N >> x;
  long long a[N];
  for (long long n = (long long)(0); n < (long long)(N); ++n) {
    cin >> a[n];
  }
  long long now = min_element(a, a + N) - a;
  42;
  long long res = 0;
  res += a[now];
  long long d[N];
  fill(d, d + N, 1000000000000000000LL);
  d[now] = a[now];
  for (long long i = (long long)(1); i < (long long)(N); ++i) {
    long long p = (now + i) % N;
    42;
    chmin(d[p], a[p]);
    chmin(d[p], d[now] + x);
  }
  42;
  cout << accumulate(d, d + N, 0LL) << endl;
  return 0;
}
long long test() { return 0; }
signed main() {
  solve();
  return 0;
}
