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>
#define ll long long
using namespace std;
 
ll gcd(ll a, ll b){
  if(b == 0){
    return a;
  }
  else{
    return gcd(b,(a%b));
  }
}
ll lcm(ll a,ll b){
  return a/gcd(a,b)*b;
}
ll mpower(ll a,ll b,ll c){
    int z;
    if(b==0){
        z=1;
        z%=c;
        return z;
    }
    if(b==1){
        z=a;
        z%=c;
        return z;
    }
    else{
        return (((mpower(a,b/2,c))*(mpower(a,b/2,c))%c)*mpower(a,b%2,c)%c);
    }
}

int main(){
  int a,b;
  cin>>a>>b;
  for(int i=1;i<1000000;i++){
    if((i*8)/100==a&&(i*10)/100==b){
      cout<<i<<endl;
      return 0;
    }
  }
  cout<<-1<<endl;
}