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 max(int a, int b) {
  return ((a > b) ? a : b);
}

int main(void) {
  int A,B;
  int pa[2],pb[2];

  /* A = 2; */
  /* B = 2; */
  /* A = 8; */
  /* B = 10; */
  /* A = 19; */
  /* B = 99; */
  scanf("%d %d", &A, &B);
  if (B < A) {
    printf("-1\n");
    return (0);
  }

  pa[0] = (int)((double)A / 0.08); // a min
  pa[1] = (int)((double)(A+1) / 0.08); // a max
  pb[0] = B * 10; // b min
  pb[1] = ((B + 1) * 10) - 1; // b max

  if ((pa[1] < pb[0]) || (pb[1] < pa[0])) {
    printf("-1\n");
    return (0);
  }

  /* printf("%d %d\n", pa[0], pb[0]); */
  /* printf("%d %d\n", pa[1], pb[1]); */

  printf("%d\n", max(pa[0], pb[0]));

  return (0);
}