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)

#pragma GCC optimize("O3")
#pragma GCC target("avx")
#pragma GCC optimize("unroll-loops")
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define DB double
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define rep(i,n) FOR(i,0,n)
//#define MOD 1000000007
#define MOD 998244353
#define INF 1145141919810364364
#define PI 3.14159265358979
#define ERR 0.00000000000001
#define NUM 200010

void swap(LL* a,LL* b){LL tmp=*a;*a=*b;*b=tmp;}
void Yn(LL a){if(a)printf("Yes\n");else printf("No\n");}
void YN(LL a){if(a)printf("YES\n");else printf("NO\n");}
LL pwmn(LL a,LL n){LL ans=1;while(ans<a)ans*=n;return ans;}
LL GCD(LL a,LL b){LL c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
LL LCM(LL a,LL b){return a*b/GCD(a,b);}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int lcmpr(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return -1;if(*(LL*)a<*(LL*)b)return 1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int ccmpr(const void* a,const void* b){return *(char*)b-*(char*)a;}
int scmp(const void* a,const void* b){return strcmp((char*)a,(char*)b);}
int scmpr(const void* a,const void* b){return strcmp((char*)b,(char*)a);}
LL mod(LL a,LL m){if(a<0)return a%m+m;else return a%m;}
LL DIV(LL a,LL d){LL m=MOD,x=1,y=0,k;while(m){k=d/m;d-=k*m;swap(&m,&d);x-=k*y;swap(&x,&y);}return mod(a*mod(x,MOD),MOD);}
LL FAC(LL a){LL i,ans=1;FOR(i,1,a+1){ans*=i;if(MOD>0&&ans>MOD)ans%=MOD;}return ans;}
LL POW(LL a,LL n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return ans;}
LL fact[NUM],finv[NUM],inv[NUM];
void comi(){LL i;fact[0]=fact[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fact[i]=fact[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
LL com(LL n,LL k){if(n<k||n<0||k<0)return 0;return fact[n]*(finv[k]*finv[n-k]%MOD)%MOD;}

int main(){
  LL A,B,ans=0,xa,na,xb,nb;
  scanf("%lld%lld",&A,&B);
  xa=A*125+124,na=A*125,xb=B*100+99,nb=B*100;
  ans=MAX(na,nb)/10;
  if(xa<nb||xb<na)ans=-1;

  printf("%lld\n",ans);
	return 0;
}