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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INI(n) scanf("%d",&n)
#define INL(n) scanf("%lld",&n)
#define INS(n) scanf("%s",s)
#define INF 10000000000
#define mod 1000000007
#define X 100000
typedef long long int ll;
ll finv[X];
ll fac[X];
ll inv[X];
ll usort(const void *a, const void *b){return *(ll*)a-*(ll*)b;}
ll dsort(const void *a, const void *b){return *(ll*)b-*(ll*)a;}
ll ssort(const void *a, const void *b){return strcmp((char*)a,(char*)b);}
ll min(ll a,ll b){
  if(a>b)return b;
  return a;
}
ll max(ll a,ll b){
  if(a>b)return a;
  return b;
}
ll z(ll a){return max(a,-a);}
ll gcd(ll a,ll b){if(!b){return a;}else{return gcd(b,a%b);}}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll kt(ll a){
  ll sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
ll ks(ll a){
  ll sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
void nCrReady(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for (ll i=2;i<X;i++){
    fac[i]=fac[i-1]*i%mod;
    inv[i]=mod-inv[mod%i]*(mod/i)%mod;
    finv[i]=finv[i-1]*inv[i]%mod;
  }
}
ll nCr(ll n,ll k){return fac[n]*(finv[k]*finv[n-k]%mod)%mod;}

int main(void){
  int a,b;
  INI(a);
  INI(b);
  FOR(101){
    if(!i)break;
    if((int)(i*0.08)==a&&(int)(i*0.1)==b){
      PRN(i);
      return 0;
    }
  }
  PRS("-1");
  return 0;
}