Problem:
Score : 300 points

Problem Statement

Find the price of a product before tax such that, when the consumption tax rate is 8 percent and 10 percent, the amount of consumption tax levied on it is A yen and B yen, respectively. (Yen is the currency of Japan.)

Here, the price before tax must be a positive integer, and the amount of consumption tax is rounded down to the nearest integer.

If multiple prices satisfy the condition, print the lowest such price; if no price satisfies the condition, print -1.

Constraints

1 \leq A \leq B \leq 100

A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

If there is a price that satisfies the condition, print an integer representing the lowest such price; otherwise, print -1.

Sample Input 1

2 2

Sample Output 1

25

If the price of a product before tax is 25 yen, the amount of consumption tax levied on it is:

When the consumption tax rate is 8 percent: \lfloor 25 \times 0.08 \rfloor = \lfloor 2 \rfloor = 2 yen.

When the consumption tax rate is 10 percent: \lfloor 25 \times 0.1 \rfloor = \lfloor 2.5 \rfloor = 2 yen.

Thus, the price of 25 yen satisfies the condition. There are other possible prices, such as 26 yen, but print the minimum such price, 25.

Sample Input 2

8 10

Sample Output 2

100

If the price of a product before tax is 100 yen, the amount of consumption tax levied on it is:

When the consumption tax rate is 8 percent: \lfloor 100 \times 0.08 \rfloor = 8 yen.

When the consumption tax rate is 10 percent: \lfloor 100 \times 0.1 \rfloor = 10 yen.

Sample Input 3

19 99

Sample Output 3

-1

There is no price before tax satisfying this condition, so print -1.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//
//  atcoder.cpp
//
//
///
#include <bits/stdc++.h>
#include <boost/optional.hpp>
#include <boost/range/irange.hpp>
#include <boost/range/algorithm.hpp>
#include <boost/range/numeric.hpp>
#include <boost/bimap/bimap.hpp>
#include <boost/bimap/multiset_of.hpp>
#include <boost/multi_array.hpp>
#include <boost/math/common_factor.hpp>
#include <boost/algorithm/string.hpp>

#define REP(i, N) for(int i=0; i<(N); i++)
#define REC(f, ...) (void)0; const auto f = [&](const auto f, __VA_ARGS__)
using namespace std;
using namespace boost::math;

using ll = long long;
namespace bst = boost;
template<typename T> class in {
    T value;
public:
    in() { cin >> value; }
    operator const T&() const { return value; }
    operator T&() { return value; }
    in& operator=(T source) { value = source; return *this; }
};
template<> class in<string> : public string {
public:
    in() { cin >> *this; }
};
template<class T> class inseq : public T {
public:
    template<typename... Arg> inseq(Arg... arg) : T(forward<Arg>(arg)...) { for (auto&& i : *this) cin >> i; }
};
template<class T> using invec = inseq<vector<T>>;

auto solve() {
  in<int> A, B;
  
  for (int i = 1; i <= 20000; i++) {
    int ka = (int)((double)i * 0.08), kb = (int)((double)i * 0.1);
    if (ka == A && kb == B)
      return i;
  }
  
  return -1;
}

template<typename T> void plot(const T& func) { cout << func() << endl; }
template<> void plot<void()>(void(&func)()) { func(); }
template<> void plot<double()>(double(&func)()) { cout << setprecision(9) << func() << endl; }

int main() {
  plot(solve);

  return 0;
}