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(void) {
	int a;//8%
	int b;//10%
	double x;//answer1 不等式の範囲を求めることで求める　下限
	double y;//answer2　上限
	double bx;
	double by;

	scanf("%d %d", &a, &b);
	x = a * 100 / 8; //これが基準　下限
	y = (a + 1) * 100 / 8; //これが上限
	bx = b * 10; //下限
	by = (b + 1) * 10;//上限
	a = (int)x;
	if((double)a == x && x < by) {
		printf("%d\n", (int)x);
		return 0;
	}
	b = (int)bx;
	if((double)b == bx && bx < y) {
		printf("%d\n", (int)bx);
		return 0;
	}
	else {
	if((int)y <= (int)bx || (int)x >= (int)by)
		puts("-1");
	else {
		if((int)bx <= (int)x) 
			printf("%d\n", (int)x);
		else
			printf("%d\n", (int)bx);
	}
	}
	return 0;
}