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(void){
	int A,B,C1,C2,D1,D2,E1,E2,i,j,ans;
	double F2,G2;
	scanf("%d%d",&A,&B);
	C1=A*100/8;
	C2=(A+1)*100/8;
	F2=(A+1.0)*100.0/8.0;
	D1=B*100/10;
	D2=(B+1)*100/10;
	G2=(B+1.0)*100.0/10.0;
	if(fmod(F2,1)!=0)C2++;
	if(fmod(G2,1)!=0)D2++;
	if(C1<D1){
		E1=C1;
		E2=C2;
		C1=D1;
		C2=D2;
		D1=E1;
		D2=E2;
	}
	for(i=D1;i<D2;i++){
		for(j=C1;j<C2;j++){
			if(i==j){
				ans=i;
				j=C2;
				i=D2;
			}
		}
	}
	if(i==D2){
		ans=-1;
	}
	printf("%d\n",ans);
	return 0;
}