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: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<int(n);i++)
#define rep1(i,n) for(int i=1;i<=int(n);i++)
#define per(i,n) for(int i=int(n)-1;i>=0;i--)
#define per1(i,n) for(int i=int(n);i>0;i--)
#define foreach(i,n) for(auto &&i:n)
#define all(x) (x).begin(),(x).end()
#define SORT(x) sort(all(x))
#define GSORT(x) sort(all(x), greater<int>())
#define REV(x) reverse(all(x))
#define MAX(x) *max_element(all(x))
#define MIN(x) *min_element(all(x))
#define LOWitr(x, n) lower_bound(all(x), n)
#define UPRitr(x, n) upper_bound(all(x), n)
#define cii(x) int x;cin >> x
#define cill(x) int64 x;cin >> x
#define cis(x) string x;cin >> x
#define co(x) cout << x << endl
#define dump(x)  cout << #x << " = " << (x) << endl
#define truecheck assert
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define fastcin() cin.tie(0);ios::sync_with_stdio(false)
using namespace std;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
typedef long long int64;
template<class T,class U>bool chmax(T &a, const U &b) { if (a<b) { a=b; return true; } return false; }
template<class T,class U>bool chmin(T &a, const U &b) { if (b<a) { a=b; return true; } return false; }
const int INF = 1e9, MOD = 1e9 + 7;
const LL LLINF = 1e16;

int main() {
    fastcin();
    cout << fixed << setprecision(10);
    cii(a);
    cii(b);
    int emin = (int)(a * 12.5);
    int emax = (int)((a+1) * 12.5 - 0.5);
    int tmin = b * 10;

    auto ans = (tmin <= emax)?max(emin, tmin):-1;

    co(ans);
    return 0;
}