stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
 #include<iostream>

 #include<algorithm>

 #include<cmath>

 #include<iomanip>

 #include<cstring>

 #include<cmath>

 #include<map>

 #include<vector>

 #include<sstream>

 #include<queue>

using namespace std;



typedef long long ll;

const ll maxn=1e18;

int main()

{

    long long sum=1;

    int n;

    long long mid;

    cin>>n;

    int flag=0;

    while(n--)

    {

        cin>>mid;

        if(mid==0)

        {

            flag=-1;

        cout<<"0";

        return 0;

        }

        if(maxn/sum>=mid)

        sum*=mid;

        else

        flag=1;

    }

    

    if(flag==1) cout<<"-1";

    else if(flag==-1)cout<<"0";

    else cout<<sum;

    

}

// Faster C++ solution for the same problem (measured speedup: 5.228800193635592x):
#include<stdio.h>



int main(){

	long long int tc;

	scanf("%lld", &tc);

	long long int n[tc], tot=1;

	for(long long int i=0; i<tc; i++){

//		tot<1000000000000000000

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

		if(n[i]==0){

			tot=0;

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

			break;

		}

	}

	if(tot>0){

	for(long long int i=tc-1; i>=0; i--){

		if(tot>1000000000 && n[i]>1000000000){

			tot=0;

			  printf("-1\n");

			break;

		}

		else if(tot*n[i]>1000000000000000000){

			tot=0;

			printf("-1\n");

		}

		else{

			tot*=n[i];	

		}

	}

	}

	if(tot>0){

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

	}

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long int n,s=1;

    double ans=0;

    cin>>n;

    long long int a[n],c=1,i;

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

        cin>>a[i];

        if(a[i]==0)

            c=0;

        //cout<<log10(a[i])<<endl;

        ans=ans+(log10(a[i]));

        //cout<<ans<<endl;

    }

    if(c==0)

        cout<<"0"<<endl;

    else if(ans<=18){

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

            s=s*a[i];

        if(s<=1000000000000000000)

        cout<<s<<endl;

        else

            cout<<"-1"<<endl;

    }

    else

        cout<<"-1"<<endl;

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.272347853601393x):
#include<stdio.h>



int main(){

	long long int tc;

	scanf("%lld", &tc);

	long long int n[tc], tot=1;

	for(long long int i=0; i<tc; i++){

//		tot<1000000000000000000

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

		if(n[i]==0){

			tot=0;

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

			break;

		}

	}

	if(tot>0){

	for(long long int i=tc-1; i>=0; i--){

		if(tot>1000000000 && n[i]>1000000000){

			tot=0;

			  printf("-1\n");

			break;

		}

		else if(tot*n[i]>1000000000000000000){

			tot=0;

			printf("-1\n");

		}

		else{

			tot*=n[i];	

		}

	}

	}

	if(tot>0){

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

	}

	return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll wq[100003];

int main()

{

    int g,t;

    ll re=1,lm=1e18;

    scanf("%d",&t);

    for(g=0;g<t;g++)

        scanf("%lld",&wq[g]);

    for(g=0;g<t;g++)if(!wq[g]){

        printf("0\n");

        return 0;

    }

    for(g=0;g<t;g++){

        if(wq[g]<=(lm/re))

            re*=wq[g];

        else{

            printf("-1\n");

            return 0;

        }

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.164941383483671x):
#include<stdio.h>



int main(){

	long long int tc;

	scanf("%lld", &tc);

	long long int n[tc], tot=1;

	for(long long int i=0; i<tc; i++){

//		tot<1000000000000000000

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

		if(n[i]==0){

			tot=0;

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

			break;

		}

	}

	if(tot>0){

	for(long long int i=tc-1; i>=0; i--){

		if(tot>1000000000 && n[i]>1000000000){

			tot=0;

			  printf("-1\n");

			break;

		}

		else if(tot*n[i]>1000000000000000000){

			tot=0;

			printf("-1\n");

		}

		else{

			tot*=n[i];	

		}

	}

	}

	if(tot>0){

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

	}

	return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define c 1000000000000000000

#define ll long long int

int main()

{

    ll n,i,count=1,flag=0;

    cin>>n;

    ll a[n+1];

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

    {

        cin>>a[i];

        if(a[i]==0)

            flag=1;

    }

    ll sum=a[0];

    if(flag==0) {

        for (i = 1; i < n; i++) {

            if (a[i] <= c / sum) {

                sum=sum*a[i];

                count++;

            } else

            {

                cout<<"-1"<<endl;

                break;

            }

        }

        if(count==n)

            cout<<sum<<endl;

    }

    else

        cout<<"0"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.225435471895266x):
#include<stdio.h>



int main(){

	long long int tc;

	scanf("%lld", &tc);

	long long int n[tc], tot=1;

	for(long long int i=0; i<tc; i++){

//		tot<1000000000000000000

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

		if(n[i]==0){

			tot=0;

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

			break;

		}

	}

	if(tot>0){

	for(long long int i=tc-1; i>=0; i--){

		if(tot>1000000000 && n[i]>1000000000){

			tot=0;

			  printf("-1\n");

			break;

		}

		else if(tot*n[i]>1000000000000000000){

			tot=0;

			printf("-1\n");

		}

		else{

			tot*=n[i];	

		}

	}

	}

	if(tot>0){

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

	}

	return 0;

}
// Slower C++ solution:
#include<iostream> 

#include<algorithm>

#include<map>

#include <cstdlib>

using namespace std;



int main(){

    int n;

    cin>>n;

    long long r=1,a[100005];

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

    	cin>>a[i];

	}

	sort(a,a+n);

	if(a[0]==0){

		cout<<0;

		return 0;

	}

	for(int i=n-1;i>=0;i--){

		if(1000000000000000000/a[i]>=r) r*=a[i];

		else{

			cout<<-1;

			return 0;

		}

	}

	cout<<r;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.313097736491035x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include<iostream>

#include<algorithm>

#include<cstring>

using namespace std;

typedef long long  ll;

const long long maxt=1e18;

int main()

{

	ll n,a[100005],b,ans=1;

	cin>>n;

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

	{

		cin>>a[i];

		if(a[i]==0)

		{

			cout<<0;

			return 0;

		}

	}

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

	{

		if((maxt/ans)<a[i])

		{

			cout<<-1;

			return 0;

		}

		ans*=a[i];

		

	}

	cout<<ans;

}

// Faster C++ solution for the same problem (measured speedup: 5.297206458194013x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

 

const int MAXN = 2e5+2;

unsigned long long nums[MAXN];

 

const long long MAXL = 1e18;

 

bool overflow(unsigned long long a, unsigned long long b) {

    if (ULLONG_MAX / a < b) {

        return true;

    }

    unsigned long long ans = a*b;

    if (ans>MAXL) {

        return true;

    }

    return false;

}

 

int main() {

    int n;

    scanf("%d", &n);

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

        scanf("%llu", &nums[i]);

        if (0==nums[i]) {

            printf("0\n");

            return 0;

        }

    }

 

    long long ans=1;

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

        if (true == overflow(ans, nums[i])) {

            printf("-1\n");

            return 0;

        }

        ans *= nums[i];

    }

 

    printf("%llu", ans);

 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.167123874017356x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main(){

	long long n, a, ans=1, p, x=1000000000000000000, flag=0;

	cin>>n;

	while(n--){

		cin>>a;

		if(ans==0)

			break;

		p=x/ans;

		if(a>p){

			ans=1;

			flag=1;

		}

		else

			ans*=a;

	}

	if(ans!=0){

		if(flag)

		cout<<-1;

		else

		cout<<ans;

	}

	else

		cout<<0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.218809753104378x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
 #include<iostream>

 #include<algorithm>

 #include<cmath>

 #include<iomanip>

 #include<cstring>

 #include<cmath>

 #include<map>

 #include<vector>

 #include<sstream>

 #include<queue>

using namespace std;



typedef long long ll;

const ll maxn=1e18;

int main()

{

    long long sum=1;

    int n;

    long long mid;

    cin>>n;

    int flag=0;

    while(n--)

    {

        cin>>mid;

        if(mid==0)

        {

            flag=-1;

        cout<<"0";

        return 0;

        }

        if(maxn/sum>=mid)

        sum*=mid;

        else

        flag=1;

    }

    

    if(flag==1) cout<<"-1";

    else if(flag==-1)cout<<"0";

    else cout<<sum;

    

}

// Faster C++ solution for the same problem (measured speedup: 5.230555267383103x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long int n,s=1;

    double ans=0;

    cin>>n;

    long long int a[n],c=1,i;

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

        cin>>a[i];

        if(a[i]==0)

            c=0;

        //cout<<log10(a[i])<<endl;

        ans=ans+(log10(a[i]));

        //cout<<ans<<endl;

    }

    if(c==0)

        cout<<"0"<<endl;

    else if(ans<=18){

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

            s=s*a[i];

        if(s<=1000000000000000000)

        cout<<s<<endl;

        else

            cout<<"-1"<<endl;

    }

    else

        cout<<"-1"<<endl;

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.274117544345526x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll wq[100003];

int main()

{

    int g,t;

    ll re=1,lm=1e18;

    scanf("%d",&t);

    for(g=0;g<t;g++)

        scanf("%lld",&wq[g]);

    for(g=0;g<t;g++)if(!wq[g]){

        printf("0\n");

        return 0;

    }

    for(g=0;g<t;g++){

        if(wq[g]<=(lm/re))

            re*=wq[g];

        else{

            printf("-1\n");

            return 0;

        }

    }

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

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.166675022692264x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

#define c 1000000000000000000

#define ll long long int

int main()

{

    ll n,i,count=1,flag=0;

    cin>>n;

    ll a[n+1];

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

    {

        cin>>a[i];

        if(a[i]==0)

            flag=1;

    }

    ll sum=a[0];

    if(flag==0) {

        for (i = 1; i < n; i++) {

            if (a[i] <= c / sum) {

                sum=sum*a[i];

                count++;

            } else

            {

                cout<<"-1"<<endl;

                break;

            }

        }

        if(count==n)

            cout<<sum<<endl;

    }

    else

        cout<<"0"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.227189416256574x):
#include<stdio.h>

int main (){



    long long int berapaKali;

    scanf("%lld", &berapaKali);

    long long int a1[berapaKali];

    long long int a2 = 1;

    long long int max = 1000000000000000000;

    long long int hasil;



    for(int j=0;j<berapaKali;j++){

        scanf("%lld", &a1[j]);

    }



    for(int h=0;h<berapaKali;h++){



        if(a1[h]==0){

            printf("0\n");

            return 0;

        }

    }

    for(int i=0;i<berapaKali;i++){

        if(a1[i] <= max/a2){

        	a2 *= a1[i]; 

		}else{

    		printf("-1\n");

			return 0;	

		}

    }

    hasil = a2;

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

    

    return 0;

}
// Slower C++ solution:
#include<iostream> 

#include<algorithm>

#include<map>

#include <cstdlib>

using namespace std;



int main(){

    int n;

    cin>>n;

    long long r=1,a[100005];

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

    	cin>>a[i];

	}

	sort(a,a+n);

	if(a[0]==0){

		cout<<0;

		return 0;

	}

	for(int i=n-1;i>=0;i--){

		if(1000000000000000000/a[i]>=r) r*=a[i];

		else{

			cout<<-1;

			return 0;

		}

	}

	cout<<r;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.32257653183643x):
#include <stdio.h>

int main (){

    long long int N;

    long long int total=1;

    

    scanf("%lld",&N);

    long long int A[N]; 

    

    for(int i=0;i<N;i++){

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

    }

    for(int j=0;j<N;j++){

        if(A[j] == 0){  

            printf("0"); 

            return 0;

        }

    }

    for(int k=0;k<N;k++){

        if(A[k]<=1000000000000000000/total){

            total = total*A[k];

            

        } else {

            printf("-1");

            return 0;

        }

    }

    printf("%lld", total);

    

    

return 0;

}
// Slower C++ solution:
#include<iostream>

#include<algorithm>

#include<cstring>

using namespace std;

typedef long long  ll;

const long long maxt=1e18;

int main()

{

	ll n,a[100005],b,ans=1;

	cin>>n;

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

	{

		cin>>a[i];

		if(a[i]==0)

		{

			cout<<0;

			return 0;

		}

	}

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

	{

		if((maxt/ans)<a[i])

		{

			cout<<-1;

			return 0;

		}

		ans*=a[i];

		

	}

	cout<<ans;

}

// Faster C++ solution for the same problem (measured speedup: 5.306656902814043x):
#include <stdio.h>

int main (){

    long long int N;

    long long int total=1;

    

    scanf("%lld",&N);

    long long int A[N]; 

    

    for(int i=0;i<N;i++){

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

    }

    for(int j=0;j<N;j++){

        if(A[j] == 0){  

            printf("0"); 

            return 0;

        }

    }

    for(int k=0;k<N;k++){

        if(A[k]<=1000000000000000000/total){

            total = total*A[k];

            

        } else {

            printf("-1");

            return 0;

        }

    }

    printf("%lld", total);

    

    

return 0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

 

const int MAXN = 2e5+2;

unsigned long long nums[MAXN];

 

const long long MAXL = 1e18;

 

bool overflow(unsigned long long a, unsigned long long b) {

    if (ULLONG_MAX / a < b) {

        return true;

    }

    unsigned long long ans = a*b;

    if (ans>MAXL) {

        return true;

    }

    return false;

}

 

int main() {

    int n;

    scanf("%d", &n);

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

        scanf("%llu", &nums[i]);

        if (0==nums[i]) {

            printf("0\n");

            return 0;

        }

    }

 

    long long ans=1;

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

        if (true == overflow(ans, nums[i])) {

            printf("-1\n");

            return 0;

        }

        ans *= nums[i];

    }

 

    printf("%llu", ans);

 

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.176342245700926x):
#include <stdio.h>

int main (){

    long long int N;

    long long int total=1;

    

    scanf("%lld",&N);

    long long int A[N]; 

    

    for(int i=0;i<N;i++){

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

    }

    for(int j=0;j<N;j++){

        if(A[j] == 0){  

            printf("0"); 

            return 0;

        }

    }

    for(int k=0;k<N;k++){

        if(A[k]<=1000000000000000000/total){

            total = total*A[k];

            

        } else {

            printf("-1");

            return 0;

        }

    }

    printf("%lld", total);

    

    

return 0;

}
// Slower C++ solution:
#include<iostream>

using namespace std;

int main(){

	long long n, a, ans=1, p, x=1000000000000000000, flag=0;

	cin>>n;

	while(n--){

		cin>>a;

		if(ans==0)

			break;

		p=x/ans;

		if(a>p){

			ans=1;

			flag=1;

		}

		else

			ans*=a;

	}

	if(ans!=0){

		if(flag)

		cout<<-1;

		else

		cout<<ans;

	}

	else

		cout<<0;

} 

// Faster C++ solution for the same problem (measured speedup: 5.228120334623794x):
#include <stdio.h>

int main (){

    long long int N;

    long long int total=1;

    

    scanf("%lld",&N);

    long long int A[N]; 

    

    for(int i=0;i<N;i++){

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

    }

    for(int j=0;j<N;j++){

        if(A[j] == 0){  

            printf("0"); 

            return 0;

        }

    }

    for(int k=0;k<N;k++){

        if(A[k]<=1000000000000000000/total){

            total = total*A[k];

            

        } else {

            printf("-1");

            return 0;

        }

    }

    printf("%lld", total);

    

    

return 0;

}
// Slower C++ solution:
 #include<iostream>

 #include<algorithm>

 #include<cmath>

 #include<iomanip>

 #include<cstring>

 #include<cmath>

 #include<map>

 #include<vector>

 #include<sstream>

 #include<queue>

using namespace std;



typedef long long ll;

const ll maxn=1e18;

int main()

{

    long long sum=1;

    int n;

    long long mid;

    cin>>n;

    int flag=0;

    while(n--)

    {

        cin>>mid;

        if(mid==0)

        {

            flag=-1;

        cout<<"0";

        return 0;

        }

        if(maxn/sum>=mid)

        sum*=mid;

        else

        flag=1;

    }

    

    if(flag==1) cout<<"-1";

    else if(flag==-1)cout<<"0";

    else cout<<sum;

    

}

// Faster C++ solution for the same problem (measured speedup: 5.239886803406199x):
#include <stdio.h>

int main (){

    long long int N;

    long long int total=1;

    

    scanf("%lld",&N);

    long long int A[N]; 

    

    for(int i=0;i<N;i++){

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

    }

    for(int j=0;j<N;j++){

        if(A[j] == 0){  

            printf("0"); 

            return 0;

        }

    }

    for(int k=0;k<N;k++){

        if(A[k]<=1000000000000000000/total){

            total = total*A[k];

            

        } else {

            printf("-1");

            return 0;

        }

    }

    printf("%lld", total);

    

    

return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    long long int n,s=1;

    double ans=0;

    cin>>n;

    long long int a[n],c=1,i;

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

        cin>>a[i];

        if(a[i]==0)

            c=0;

        //cout<<log10(a[i])<<endl;

        ans=ans+(log10(a[i]));

        //cout<<ans<<endl;

    }

    if(c==0)

        cout<<"0"<<endl;

    else if(ans<=18){

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

            s=s*a[i];

        if(s<=1000000000000000000)

        cout<<s<<endl;

        else

            cout<<"-1"<<endl;

    }

    else

        cout<<"-1"<<endl;

    return 0;

}




// Faster C++ solution for the same problem (measured speedup: 5.283526797348167x):
#include <stdio.h>

int main (){

    long long int N;

    long long int total=1;

    

    scanf("%lld",&N);

    long long int A[N]; 

    

    for(int i=0;i<N;i++){

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

    }

    for(int j=0;j<N;j++){

        if(A[j] == 0){  

            printf("0"); 

            return 0;

        }

    }

    for(int k=0;k<N;k++){

        if(A[k]<=1000000000000000000/total){

            total = total*A[k];

            

        } else {

            printf("-1");

            return 0;

        }

    }

    printf("%lld", total);

    

    

return 0;

}