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)

#include <bits/stdc++.h>
using namespace std;

constexpr int64_t INF = 1'010'000'000'000'000'017LL;
constexpr int64_t MOD = 1'000'000'007LL;
constexpr double EPS = 1e-12;
constexpr double PI = 3.14159265358979323846;

#define FOR(i, start, end) for(int64_t i=start; i<end; i++)
#define REP(i, n) FOR(i, 0, n)

// 最大公約数gcd
uint64_t gcd(uint64_t m, uint64_t n) {
  uint64_t temp;
  while (m % n != 0){
    temp = n;
    n = m % n;
    m = temp;
  }
  return n;
}

// 最小公倍数lcm=m*n/gcd
uint64_t lcm(uint64_t m, uint64_t n) {
  return (m*n)/gcd(m,n);
}

// nCm
void comb(vector<vector <uint64_t> > &v){
  for(uint64_t i = 0;i <v.size(); i++){
    v[i][0]=1;
    v[i][i]=1;
  }
  for(uint64_t k = 1;k <v.size();k++){
    for(uint64_t j = 1;j<k;j++){
      v[k][j]=(v[k-1][j-1]+v[k-1][j]);
    }
  }
}

signed main() {
  int64_t a, b;
  cin >> a >> b;
  
  int x1min = ((float)a) * 12.5f + 0.5f;
  int x1max = ((float)a) * 12.5f + 1.0f/0.08f - 1.0f;
  
  int x2min = b * 10;
  int x2max = b * 10 + 9;
  
  if ( x2min <= x1min && x1min <= x2max ) {
  	cout << x1min << endl;
  }
  else if ( x1min <= x2min && x2min <= x1max ) {
  	cout << x2min << endl;
  }
  else{
    cout << "-1" << endl;
  }
  
  return 0;
}