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: Wrong Answer)

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

typedef long long ll;

int main()
{
    int a,b;
    cin >> a >> b;
    int i;
    vector<int> temp_a(2000);
    for(i = 0;i < 2000;++i){
        double ti = i;
        temp_a.at(i) = (int)(ti*1.08) - i;
    }
    vector<int> temp_b(2000);
    for(i = 0;i < 2000;++i){
        double ti = i;
        temp_b.at(i) = (int)(ti*1.1) - i;
    }

    int id_a = lower_bound(temp_a.begin(), temp_a.end(), a) - temp_a.begin();
    int id_a1 = upper_bound(temp_a.begin(), temp_a.end(), a) - temp_a.begin();
    int id_b = lower_bound(temp_b.begin(), temp_b.end(), b) - temp_b.begin();
    int id_b1 = upper_bound(temp_b.begin(), temp_b.end(), b) - temp_b.begin();

    if(id_a > id_b1 || id_b > id_a1){
        cout << -1 << endl;
    }else{
        cout << max(id_a, id_b) << endl;
    }

    return 0;
}