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)

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          | | :  `- \`. ;`. _/; .'/ /  .' ; |
          \  \ `-.   \_\_`. _.'_/_/  -' _.' /
===========`-.`___`-.__\ \___  /__.-'_.'_.-'================

                  Please give me AC.
*/

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
#include <complex>
#include <bitset>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <chrono>
#include <random>

using namespace std;

using int64 = long long;
using uint64 = unsigned long long;
using vi = vector<int>;
using vl = vector<int64>;
using pii = pair<int, int>;
using pll = pair<int64, int64>;

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(v) (v).begin(), (v).end()
#define print(x) cout << (x) << '\n'
#define print2(x, y) cout << (x) << ' ' << (y) << '\n'
#define print3(x, y, z) cout << (x) << ' ' << (y) << ' ' << (z) << '\n'
#define printn(v) rep(i, (v).size() - 1) cout << (v)[i] << ' '; cout << (v)[n - 1] << '\n';
#define debug(x) cerr << #x << ": " << (x) << '\n'
#define debug2(x, y) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << '\n'
#define debug3(x, y, z) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << ", " << #z << ": " << (z) << '\n'
#define dbg(v) for (size_t _ = 0; _ < v.size(); ++_){cerr << #v << "[" << _ << "] : " << v[_] << '\n';}
// constant
const int INF = (1<<30) - 1;
const int64 INF64 = (1LL<<62) - 1;

template<typename T> T gcd(T a, T b) {
    if (a < b) return gcd(b, a);
    T r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

template<typename T> T lcm(const T a, const T b) {
    return a / gcd(a, b) * b;
}

template<typename T> bool chmin(T& a, const T& b) {
    if (a > b) return a = b, true; else return false;
}

template<typename T> bool chmax(T& a, const T& b) {
    if (a < b) return a = b, true; else return false;
}

// End of template.

int main() {
    cout << fixed << setprecision(15);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    // cin.read(buf, sizeof buf); // 注意: ./a.out < in か pbp | ./a.out で入力すること

    int a, b;
    cin >> a >> b;
    rep(i, 5000) {
        if (8 * i / 100 == a && 10 * i / 100 == b) {
            print(i);
            return 0;
        }
    }

    print(-1);

    return 0;
}