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>
#include <stdlib.h>
#include <string.h>

int main()
{
    float a,b;
    scanf("%f %f",&a,&b);
    float min1,min2,max1,max2;
    int i1,i2,x1,x2;
    min1=a*100/8;
    min2=b*100/10;
    max1=min1+12.5;
    max2=min2+10;
    i1=(int)min1;
    i2=(int)min2;
    x1=(int)max1;
    x2=(int)max2;
    if((i1<i2&&x1<i2)||(i2<i1&&x2<i1))
    {
        printf("-1\n");
    }
    else if((i1<0&&i2<0))
    {
        printf("-1\n");
    }
    else
    {
        if(i1<i2)
        {
            if((int)(i2*10/100)!=(int)b)
            {
                printf("%d\n",i2+1);
            }
            else
            {
                printf("%d\n",i2);
            }
        }
        else
        {
            if((int)(i1*8/100)!=(int)a)
            {
                printf("%d\n",i1+1);
            }
            else
            {
                printf("%d\n",i1);
            }
        }
    }

    return 0;
}