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;
using ll = long long;
using vi = vector<int>;
void ins() {}
template <class T, class... Rest>
void ins(T& v, Rest&... rest) {
  cin >> v;
  ins(rest...);
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
  os << "{";
  for (size_t i = 0; i < vec.size(); ++i) os << (i ? ", " : "") << vec[i];
  os << "}";
  return os;
}
ostream& operator<<(ostream& os, const vector<char>& v) {
  for (size_t i = 0; i < v.size(); ++i) os << v[i];
  return os;
}
template <class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& rhs) {
  os << "(" << rhs.first << ", " << rhs.second << ")";
  return os;
}
int main() {
  int N;
  ll x;
  ins(N, x);
  vector<ll> A(N);
  for (int i = 0, _i = (N); i < _i; ++i) cin >> A[i];
  vector<bool> bit(N, true);
  ll D = 0, ans = 0;
  for (int k = 0, _i = (N); k < _i; ++k) {
    for (int i = 0, _i = (N); i < _i; ++i) {
      int idx = (i - k + N) % N;
      if (bit[idx] && A[i] <= D + x) {
        bit[idx] = false;
        ans += A[i];
      }
    }
    bool ok = true;
    for (int i = 0, _i = (N); i < _i; ++i)
      if (bit[i]) ok = false;
    if (ok) {
      cout << ans + D << endl;
      return 0;
    }
    D += x;
  }
  return 0;
}
