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>
//#include <ext/pb_ds/assoc_container.hpp> 
//#include <ext/pb_ds/tree_policy.hpp>
//#include <functional> // for less
//using namespace __gnu_pbds;

#define max_(x, y) (((x) > (y)) ? (x) : (y))
#define min_(x, y) (((x) < (y)) ? (x) : (y))
#define ___ ios_base::sync_with_stdio(false);cin.tie(NULL);
using namespace std;
#define Mx 200001
#define inf 1e15
const int MOD = 1e9+7;
const double pi = 2 * acos (0.0);

/*------------------------------Graph Moves----------------------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*---------------------------------------------------------------------*/


typedef long long  L;
//typedef tree<L,null_type,less<L>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;

L a,b,a1,a2,a3,b1,b2,b3,c,c1,c2,c3,d,d1,d2,d3,d4,d5,e1,e2,e3,f,h,k,l,m,n,m1,m2,n1,n2,o,p,p1,p2,p3,p4,q,q1,q2,q3,q4,r,s,t,
u,v,x,y,z,sum,mx=-inf,mn=inf,flg,flg1,flg2,id1,id2,id3,id4,lim;

//double rat;
L A[Mx+5],B[Mx+5];


//struct node{L nid,nv;node(L nid_,L nv_){nid=nid_;nv=nv_;}};
//vector<node>V;

void solve(){
	d1=(a*100)/8;
	d2=(b*10);
	if(d1>d2)swap(d1,d2);
	for(L i=d1;i<=d2;i++){
		c1=i*8/100;
		c2=i/10;
		if(c1==a&&c2==b){
			printf("%lld\n",i);return;
		}
	}
	printf("-1\n");
}
int main()
{
    scanf("%lld%lld",&a,&b);
    solve();
    return 0;
}
/// lld I64d push_back() top() pop() insert() upper_bound() lower_bound() memset() max() min() scanf continue