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(){
  int a,b,ans=0,cnt=0,s,t;
  scanf("%d%d",&a,&b);
  double x,y,i;
  x=a/0.08;
  s=x*10;
  if(s%10!=0){
    x=x+0.5;
  }
  y=b/0.1;
  t=y*10;
  if(t%10!=0){
    y=y+0.5;
  }
  if(x>y){
    for(i=0; i<1; i+=0.1){
      y=(b+i)/0.1;
      if(x==y){
        ans=x;
        cnt++;
        break;
      }
    }
    if(cnt==0){
      ans=-1;
    }
  }
  else{
    for(i=0; i<1; i+=0.1){
      x=(a+i)/0.08;
      if(x==y){
        ans=y;
        cnt++;
        break;
      }
    }
    if(cnt==0){
      ans=-1;
    }
  }
  printf("%d\n",ans);
  return 0;
}