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)

#include<bits/stdc++.h>
using namespace std;
//#include<boost/multiprecision/cpp_int.hpp>
//using boost::multiprecision::cpp_int;

#define int long long
#define cint cpp_int
#define endl "\n"
#define REP(i,a,n) for(int i=a;i<n;++i)
#define REPR(i,a,n) for(int i=a;i>n;--i)
#define RUP(a,b) (((a)+(b)-1)/(b))
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define MOD 1000000007
#define INF LLONG_MAX/2
#define PI acos(-1.0)

typedef long long ll;
typedef pair<int,int> Pii;
typedef tuple<int,int,int> Tiii;
typedef vector<int> Vi;
typedef vector<Vi> VVi;
typedef vector<Pii> VPii;
typedef vector<string> Vs;
typedef priority_queue<int> PQi;

template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
void vout(Vi &v){for(int i=0;i<(v).size();i++) cout<<v[i]<<" ";cout<<endl;}
void vout(Vs &v){for(int i=0;i<(v).size();i++) cout<<v[i]<<" ";cout<<endl;}
void vvout(VVi &v){for(int i=0;i<(v).size();i++) vout(v[i]);}
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
int mypow(int x, int n, int m){if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int digit(int k,int i){string s = to_string(k);return s[s.size()-i]-'0';}
void uniq(Vi &v){sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end());} //vectorの同じ要素消す
int ctoi(char c){if(c>='0'&&c<='9'){return c-'0';}return 0;}
int leng(int n){string s=to_string(n); return s.size();} //intの桁数
int digitsum(int n){int ret=0; while(n>0){ret+=n%10;n/=10;}return ret;} //桁和
int lesscount(int x,Vi &a){return lower_bound(a.begin(),a.end(),x)-a.begin();} //vector aにおけるx未満の数の個数
int orlesscount(int x,Vi &a){return upper_bound(a.begin(),a.end(),x)-a.begin();} //vector aにおけるx以下の数の個数
int morecount(int x,Vi &a){return a.size()-orlesscount(x,a);} //vector aにおけるxより大きい数の個数
int ormorecount(int x,Vi &a){return a.size()-lesscount(x,a);} //vector aにおけるx以上の数の個数
int count(int x,Vi &a) {return upper_bound(ALL(a),x)-lower_bound(ALL(a),x);} //vector aにおけるxの個数
Vi accum(Vi &v){Vi ret((v).size()+1);REP(i,0,(v).size()) ret[i+1]=ret[i]+v[i];return ret;}
bool comp(Pii a,Pii b){
    if(a.second != b.second) return a.second<b.second;
    else return a.first<b.first;
}


void solve(long long A, long long B){
    REP(i,1,1000000){
        if(i*8/100==A && i*10/100==B){
            cout<<i<<endl;
            return;
        }
    }
    cout<<-1<<endl;
    return;
}

// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
signed main(){cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);
    long long A;
    scanf("%lld",&A);
    long long B;
    scanf("%lld",&B);
    solve(A, B);
    return 0;
}