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 <math.h>

int main(){
  int ans = -1;
  double a, b, x, y;
  
  scanf("%lf %lf", &a, &b);
  x = floor(a / 0.08);
  y = floor(b / 0.1);
  
  //printf("%f %f\n", x, y);
  //printf("%d %d\n", (int)(x * 0.08) , (int)(x * 0.1));
  if(x == y){
    ans = (int)x;
  }
  else if(x > y){
    if((int)(x * 0.08) == (int)(x * 0.1)){
      ans = (int)x;
    }
    else{
      while((int)(x * 0.1) == (int)((x + 1) * 0.1)){
        x++;
        if((int)(x * 0.08) == (int)(x * 0.1)){
      	  ans = (int)x;
          break;
    	}
      }
    }
  }
  
  printf("%d\n", ans);
  
  return 0;
}