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++14 (GCC 5.4.1) (judge verdict: Accepted)

//author:luckYrat(twitter:@luckYrat_)
//#include <bits/stdc++.h>

//def
#include <iostream>

#include <cmath>
#include <algorithm>
#include <iomanip>

//array
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <utility>
#include <climits>



using namespace std;

using ll = long long;
using P = pair<ll,ll>;

#define anyfill(n,s) setw(n) << setfill(s)
#define loop(s) for(int i = 0; s > i; i++)
#define rep(i,q) for(int i = 0; (q) > i; i++)
#define repp(i,n,q) for(int i = n; (q) > i; i++)
#define dep(i,q) for(int i = (q); 0 < i; i--)


#define MAX 1000000000
const int mod = 1000000007;
#define EPS (1e-10)

#define pb push_back
#define fir first
#define scn second
#define ednl endl

#define YesNo(a) (a?"Yes":"No")
#define YESNO(a) (a?"YES":"NO")
#define yesno(a) (a?"yes":"no")


P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}};
P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
/*
確認ポイント
cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる
vector.unique()は隣接した同じ要素を削除するので先にソートをする

計算量は変わらないが楽できるシリーズ
C++11以降でmin({a,b,c...})で複数個のmic,maxを処理できる
min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる
count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる

*/
__attribute__((constructor))
void initial() {
 cin.tie(0);
 ios::sync_with_stdio(false);
}
int main(){
  int a,b;cin>>a>>b;
  for(int i = 0; 100000 > i; i++){
    if(i*8/100 == a && i*10/100 == b){
      cout << i << endl;
      return 0;
    }
  }
  cout << -1 << endl;

}