// Slower C++ solution:
//            ,O,i=0,Q=YUKU(f);static f*P;static FILE*t;static cons  char*o[]={//bad"

//          "\n\40\"8oCan\40not\40open %s\n\0\n\40\"8oCan\40not\40open %s\n#e(bad/bad",

//        "a6if(p==b*5&&O){p-=b;O--;}}return 0U;if(p==b*5&&O){p-=b;O--;}}return 0U;:b"",

//       "e&a6if(p==b*5&&O){p-=b;O--;}}return 0U;if(p==b*5&&O){p-=b;O--;}}return 0U;:b"",

//      "e&a6if(p==b*5&&O){p-=b;O--;}}return 0U;if(p==b*5&&O){p-=b;O--;}}return 0U;:"e#d",

//     ">bad/bad/"};static ll  S(){return(o[p][q]);}static/**/ll /**/Z=0  ;void/**/z(ll //

//    l){if(/**/Z-l){Z=l;q++;if(p<b*5&&!S()){p+=b;q=0;}}}ll  main(ll  I,    /**/char**l){//

//   d=YUKKUU(f*);if(1<(O=_)){b=((YUKKUU(o)/YUKKUU(char*))-1)/4;q=22; p=     0;while(p<b*5){

//  /*<*/if(Z-1){d=S()>96;i=S()-(d?96:32) ;q++;if(p<b*5&&!S()){p+=b;  q=      0;}Z=1;}/*[[*/

//  while(i){_=o[0][S()-97];I=_-10?b:1;   for( ;I--;)putchar(_ );if   (!      --i||d)z(~i );}

// if(p==b*5&&O){p-=b;O--;}}return 0U;   }if(! (P=( f*)calloc /*]*/  (Q        ,I)))return 1;

// {;}for(_=p=1;p<I;p++){e(q=1);while    (q<   p&&  strcmp(  l[p     ]         ,l[(q)]))++  q;

// t=stdin;if(q<p){(void)memcpy/* "      */    (&P  [p],&P   [q     ]          ,Q);continue ;}

//if(strcmp(l[p],"-")){t=fopen(l         [     p]   ,"rb"   )                  ;if(!t ){{;}  ;

//prll f(05+*o,l[p ]);return+1;                      {;}                       }}_=b= 1<<16   ;

//*&O=5;do{if(!(P[p].q=realloc   (P[p].q,(P[p].P     +=       b)+1))){return   01;}O   &=72   /

//6/*][*/;P[p].o+=d=fread(P[p]      .q       +P[     p           ].       o,  1,b,t)   ;}//

// while(d==b)      ;P [p].q[       P[       p]                  .o       ]=  012;d    =0;

// e(fclose(t        )  );P         [p]      .p                  =P[      p]  .q;if    (O)

// {for(;d<P[            p]          .o     ;d=                   q+     1)    {q=     d;

//  while(q<P[                        p].o&&P[                    p].q[q]-     10     ){

//  q++;}b=q-d;                         _=P                         [p].        d     ;

//  if(b>_){/*]b                                                                */

//   P[p].d=b;}{;                                                                }

//   #undef/*pqdz'.*/  e//                                                      ;

//   #define/*s8qdb]*/e/**/0                                                   //

//   //<<.<<.----.>.<<.>++.++<                                              .[>]

//   /*P[*/P[p].b++;continue;}}}t=                                       Yuki;

//  for (p=1;p<I;p++){/**/if(P[p].b>i                               ){i=P[p].b;}}

// if  (O){for(p=0;p<i;p++){q=0;/*[*/while(I               >++q){_=P[q].p-P[q ].q;

//b=   0;if(_<P[q ].o){while(012-*P[q].p)     {putchar(*(P[q].p++));b++;}P[q]. p++;

//}   ;while (P[  q].d>b++)YUKKUU(040);}             putchar(10);}return 0;}p   =1;

//   for(;   p<I   ;p++)YUKKUU(P[p] .q,P[              p].o,1,t);return 0 ;}//

//  #/*]     ]<.    [-]<[-]<[- ]<[    -]<               [-  ]<;*/Yuki  e    //b

// |(1        <<     ( __LINE__        /*               >>   `*//45))  |     01U

//

#include<bits/stdc++.h>

using namespace std;

long long int n,a,f,f2;

vector<long double>v;

long double b;

int main()

{

    cin>>n;

    b=1,f=0,f2=0;

    for(int i=0; i<n; i++)

    {

        cin>>a;

        if(a==0)

            f=1;

        else if(a>1000000000000000000)

            f2=1;

        v.push_back(a);

    }

    if(f)

        return cout<<0<<endl,0;

    if(f2)

        return cout<<-1<<endl,0;

    for(int i=0; i<n; i++)

    {

        b*=v[i];

       // cout<<b<<" ";

        if(b>1000000000000000000)

            return cout<<-1<<endl,0;

    }

    cout<<fixed<<setprecision(0)<<b;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.863532625351649x):
#include <stdio.h>



int main(){



    long long int angka;

    long long int total = 1;

    long long int N[100005];

	bool lanjut = true;

	bool printtotal = true;

    scanf("%lld", &angka);

    //printf("test");

    

    //angka apapun yang dikali 0 akan menjadi 0

    

    for (long long int i = 1; i <= angka; i++){

        scanf("%lld", &N[i]);

        if(N[i]==0){

        	printf("0\n");

			lanjut = false;

        	break;

		}

	}

	

  	if(lanjut){

  		for(int i = 1; i<=angka;i++){

  			//printf("%lld\n",total);

  			if(1000000000000000000/total >= N[i]){

  				total *= N[i];

			}

			else{

				printf("-1\n");

				printtotal = false;

				break;

			}	

		}	

		if(printtotal){

			printf("%lld\n",total);

		}

	}

    return 0;

}