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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include"stdio.h"
int main()
{
    float i,j,a1,a2,a3,a4,b1,b2,k,t;
    int x,y,tag=1;
    scanf("%f %f",&i,&j);
    a1=i*25/2;
    a2=j*10;
    x=(int)(a1+0.5)>(int)a1?(int)a1+1:(int)a1;
    y=(int)(a2+0.5)>(int)a2?(int)a2+1:(int)a2;

    if(x<y)
    {
        k=y*0.08;
        t=y*0.1;
        if(k-t>=1 || t-k>=1)
            tag=0;
    }
    else if(x>y)
    {
        k=x*0.08;
        t=x*0.1;
        if(k-t>=1 || t-k>=1)
            tag=0;
    }
    if(tag==0)
        printf("-1");
    else if(x<y && tag==1)
        printf("%d",y);
    else if(x>=y && tag==1)
        printf("%d",x);
    return 0;
}