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 rep(i,n) for(int i=0; i<(n); i++)
#define reps(i,s,n) for(int i=(s); i<(n); i++)
#define all(v) v.begin(),v.end()
#define outve(v) for(auto i : v) cout << i << " ";cout << endl
#define outmat(v) for(auto i : v){for(auto j : i) cout << j << " ";cout << endl;}
#define in(n,v) for(int i=0; i<(n); i++){cin >> v[i];}
#define out(n) cout << (n) << endl
#define fi first
#define se second
#define pb push_back
#define mod 1000000007
#define infi 1900000000
#define infl 1100000000000000000
#define cyes cout << "Yes" << endl
#define cno cout << "No" << endl
#define csp << " " <<
#define outset(n) cout << fixed << setprecision(n);
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
template<typename T> using ve = vector<T>;
template<typename T> using pq2 = priority_queue<T>;
template<typename T> using pq1 = priority_queue<T,vector<T>,greater<T>>;




int main()
{

    int A,B;
    cin >> A >> B;
    
    rep(i,1000000){
        double a = (double)i * 0.08;
        double b = (double)i * 0.1;
        if((int)a == A && (int)b == B){
            cout << i << endl;
            return 0;
        }
    }
    cout << -1 << endl;
    
    

    return 0;
}