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;
bool __hack = std::ios::sync_with_stdio(false);
auto __hack1 = cin.tie(nullptr);
namespace template_util {
constexpr int bytecount(uint64_t x) { return x ? 1 + bytecount(x >> 8) : 0; }
template <int N>
struct bytetype {};
template <>
struct bytetype<4> {};
template <>
struct bytetype<3> {};
template <>
struct bytetype<2> {};
template <>
struct bytetype<1> {};
template <>
struct bytetype<0> {};
template <uint64_t N>
struct minimal_uint : bytetype<bytecount(N)> {};
}  // namespace template_util
template <class T>
T next(istream& in) {
  T ret;
  in >> ret;
  return ret;
}
template <class T>
vector<T> next_vector(istream& in, size_t n) {
  vector<T> ret(n);
  for (size_t i = 0; i < n; ++i) {
    ret[i] = next<T>(in);
  }
  return ret;
}
void solve(istream& in, ostream& out) {
  int n = next<int>(in);
  int X = next<int>(in);
  vector<int> a = next_vector<int>(in, n);
  int64_t res = 0;
  vector<int> has(n, 0);
  while (true) {
    bool ok = true;
    for (int i = 0; i < (n); ++i) {
      if (!has[i]) {
        ok = false;
        break;
      }
    }
    if (ok) {
      out << res << endl;
      return;
    }
    int id = -1;
    for (int i = 0; i < (n); ++i) {
      if (a[i] <= X && !has[i]) {
        id = i;
        break;
      }
    }
    if (id == -1) {
      has.insert(has.begin(), 0);
      has[0] = has.back();
      has.pop_back();
      res += X;
      continue;
    }
    res += a[id];
    has[id] = 1;
  }
}
int main() {
  solve(cin, cout);
  return 0;
}
