stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



using ll = long long;

using ull = unsigned long long;

using ld = long double;



#define MOD 1000000007LL



#define rep(i, n) for(ll (i) = 0LL;(i) < (ll)(n);(i)++)

#define rep2(i, s, e) for(ll (i) = (ll)(s);(i) < (ll)(e);(i)++)

#define repi(i, n) for(ll (i) = 0LL;(i) <= (ll)(n);(i)++)

#define repi2(i, s, e) for(ll (i) = (ll)(s);(i) <= (ll)(e);(i)++)

#define per(i, n) for(ll (i) = (ll)(n) - 1LL;(i) >= 0LL;(i)--)

#define per2(i, s, e) for(ll (i) = (ll)(s) - 1LL;(i) >= (ll)(e);(i)--)

#define peri(i, n) for(ll (i) = (ll)(n);(i) >= 0LL;(i)--)

#define peri2(i, s, e) for(ll (i) = (ll)(s);(i) >= (ll)(e);(i)--)

#define iter(i, it) for(auto &(i): (it))



template<typename T, typename U> ostream& operator<<(ostream &s, const pair<T, U> m) {

    cout << "(" << m.first << ", " << m.second << ")";

    return s;

}

template<typename T, typename U> ostream& operator<<(ostream &s, const map<T, U> m) {

    ll c = 0;

    cout << "{ ";

    iter(i, m) cout << i << (c++ == m.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}

template<typename T> ostream& operator<<(ostream &s, const vector<T> &v) {

    cout << "{ ";

    rep(i, v.size()) cout << v[i] << (i == v.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}

template<typename T> ostream& operator<<(ostream &s, const list<T> &v) {

    ll c = 0;

    cout << "{ ";

    iter(i, v) cout << i << (c++ == v.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}



int main(void) {

    ll N;

    ll ans = 1;

    bool flag = false;

    bool zero = false;

    cin >> N;

    rep(i, N) {

        ll prev;

        ll temp;

        cin >> temp;

        prev = ans;

        ans *= temp;

        if(prev > ans || ans > 1000000000000000000 || temp > 1000000000000000000 || (temp > 1 && temp >= 1000000000000000000)) {

            flag = true;

        }

        if(temp == 0) zero = true;

    }

    if(zero) cout << 0 << endl;

    else if(flag) cout << -1 << endl;

    else cout << ans << endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.669211299765641x):
#define _CRT_SECURE_NO_WARNINGS

#include<stdio.h>



int main(void){

    int i;

    int N;

    long m = 1;

    long a[100000];

    int ng = 0;

    int zero = 0;



    scanf("%d", &N);



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

        long prv;

        scanf("%ld", &a[i]);

        prv = m;

        m *= a[i];

        if(prv > m || m > 1000000000000000000) ng = 1;

        if(a[i] == 0) zero = 1;

    }



    if(zero) puts("0");

    else if(ng) puts("-1");

    else printf("%ld", m);



    return 0;

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

using namespace std;



using ll = long long;

using ull = unsigned long long;

using ld = long double;



#define MOD 1000000007LL



#define rep(i, n) for(ll (i) = 0LL;(i) < (ll)(n);(i)++)

#define rep2(i, s, e) for(ll (i) = (ll)(s);(i) < (ll)(e);(i)++)

#define repi(i, n) for(ll (i) = 0LL;(i) <= (ll)(n);(i)++)

#define repi2(i, s, e) for(ll (i) = (ll)(s);(i) <= (ll)(e);(i)++)

#define per(i, n) for(ll (i) = (ll)(n) - 1LL;(i) >= 0LL;(i)--)

#define per2(i, s, e) for(ll (i) = (ll)(s) - 1LL;(i) >= (ll)(e);(i)--)

#define peri(i, n) for(ll (i) = (ll)(n);(i) >= 0LL;(i)--)

#define peri2(i, s, e) for(ll (i) = (ll)(s);(i) >= (ll)(e);(i)--)

#define iter(i, it) for(auto &(i): (it))



template<typename T, typename U> ostream& operator<<(ostream &s, const pair<T, U> m) {

    cout << "(" << m.first << ", " << m.second << ")";

    return s;

}

template<typename T, typename U> ostream& operator<<(ostream &s, const map<T, U> m) {

    ll c = 0;

    cout << "{ ";

    iter(i, m) cout << i << (c++ == m.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}

template<typename T> ostream& operator<<(ostream &s, const vector<T> &v) {

    cout << "{ ";

    rep(i, v.size()) cout << v[i] << (i == v.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}

template<typename T> ostream& operator<<(ostream &s, const list<T> &v) {

    ll c = 0;

    cout << "{ ";

    iter(i, v) cout << i << (c++ == v.size() - 1 ? " " : ", ");

    cout << "}";

    return s;

}



int main(void) {

    ll N;

    ll ans = 1;

    bool flag = false;

    bool zero = false;

    cin >> N;

    rep(i, N) {

        ll prev;

        ll temp;

        cin >> temp;

        prev = ans;

        ans *= temp;

        if(prev > ans || ans > 1000000000000000000) {

            flag = true;

        }

        if(temp == 0) zero = true;

    }

    if(zero) cout << 0 << endl;

    else if(flag) cout << -1 << endl;

    else cout << ans << endl;

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 4.669234657990119x):
#define _CRT_SECURE_NO_WARNINGS

#include<stdio.h>



int main(void){

    int i;

    int N;

    long m = 1;

    long a[100000];

    int ng = 0;

    int zero = 0;



    scanf("%d", &N);



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

        long prv;

        scanf("%ld", &a[i]);

        prv = m;

        m *= a[i];

        if(prv > m || m > 1000000000000000000) ng = 1;

        if(a[i] == 0) zero = 1;

    }



    if(zero) puts("0");

    else if(ng) puts("-1");

    else printf("%ld", m);



    return 0;

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

using namespace std;

int main()

{

	long long n;

	cin>>n;

	long long a[n];

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

	cin>>a[i];

	long long sum=1,jilu=0;

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

	{

		if(a[i]==0)

		jilu=1;

	}

	if(jilu==1)

	cout<<0;

	else

	{

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

	{

		if((1000000000000000000/a[i])<sum)

		{

			sum=-1;

			break;

		}

	sum*=a[i];

	}

	cout<<sum;

		}

} 

// Faster C++ solution for the same problem (measured speedup: 5.219512585064711x):
#include <stdio.h>



int main(){

 	long long N;

 	scanf("%lld", &N);

 	long long A[N], sum = 1;

 	

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

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

	}

	

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

 		if(A[i] == 0){

 			printf("0\n");

 			return 0;

		 }

	}

	

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

		if(A[i] <= (1000000000000000000/sum)){

			sum *= A[i];

 		} else{

 			printf("-1\n");

            return 0;

		 }

 	}



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

 	

	return 0;

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

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

#define ll  long long

#define ull unsigned long long

#define IO ios::sync_with_stdio(false), cin.tie(0)

#define endl '\n'

#define main() int main()

#define ff1(i,n) for(int i=1;i<=n;++i)

#define ff0(i,n) for(int i=0;i<n;++i)

#define pi acos(-1.0)

#define ms(a, x) memset(a, x, sizeof(a))

#define lowbit(x) ((x)&(-(x)))

#define pb push_back

#define binarycount __builtin_popcount(x)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define fir first

#define sec second

#define debug(x) cout<<x<<'\n'

#define sa(a) scanf("%d", &a)

#define sab(a, b) scanf("%d%d", &a, &b)

#define pa(a) printf("%d",a)

#define pab(a,b) printf("%d%d",a,b)

#define pii pair<int,int>

using namespace std;



#define maxn 200005

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

const double eps = 1e-8;

ll gcd(ll a, ll b) {return b==0?a:gcd(b, a%b);}

ll lcm(ll a, ll b) {return a/gcd(a, b)*b;}

ll powmod(ll a,ll b){

    ll ans = 1, pow = a%mod;

    while(b) { if(b&1) ans = (ans*pow)%mod; pow = (pow*pow)%mod; b >>= 1; }

    return ans%mod;

}

ull a[maxn];

const ll maxx=1e18;

bool judge(ull a, ull b) {

    if (maxx/a<b) {

        return true;

    }

    return false;

}

main() {

    int n;

    cin>>n;

    int f=1;

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

        cin>>a[i];

        if (a[i]==0) f=0;

    }

    if(f==0){

    	cout<<0<<endl;

    	return 0;

	}

    ll ans=1;

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

        if (judge(ans,a[i])) {

            cout<<"-1"<<endl;

            return 0;

        }

        ans*=a[i];

    }

    cout<<ans<<endl;





}

 

// Faster C++ solution for the same problem (measured speedup: 5.237071537920072x):
#include <stdio.h>



int main(){

 	long long N;

 	scanf("%lld", &N);

 	long long A[N], sum = 1;

 	

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

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

	}

	

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

 		if(A[i] == 0){

 			printf("0\n");

 			return 0;

		 }

	}

	

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

		if(A[i] <= (1000000000000000000/sum)){

			sum *= A[i];

 		} else{

 			printf("-1\n");

            return 0;

		 }

 	}



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

 	

	return 0;

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

#define IO ios::sync_with_stdio(0), cin.tie(0)

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 5;

const ll inf = 1e18;

int main() {

    IO;

    int n;

    ll ans = 1;

    cin >> n;

    bool flag = true;

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

        ll x;

        cin >> x;

        if (x == 0) {

            ans = 0;

            flag = true;

        }

        else if (ULLONG_MAX / ans < x || x * ans > inf) {

            flag = false;

            //cout << ans << " " << ULLONG_MAX << " " << ULLONG_MAX / ans << endl;

        }

        if (flag) ans *= x;

        // cout << ans << endl;

    }

    if (flag)

        cout << ans << endl;

    else {

        cout << -1 << endl;

    }



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.303214299401932x):
#include <stdio.h>



int main(){

 	long long N;

 	scanf("%lld", &N);

 	long long A[N], sum = 1;

 	

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

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

	}

	

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

 		if(A[i] == 0){

 			printf("0\n");

 			return 0;

		 }

	}

	

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

		if(A[i] <= (1000000000000000000/sum)){

			sum *= A[i];

 		} else{

 			printf("-1\n");

            return 0;

		 }

 	}



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

 	

	return 0;

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

using namespace std;

long long int n,a,b,f;

vector<long long int>v;

int main()

{

    cin>>n;

    b=1,f=0;

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

    {

        cin>>a;

        if(a==0)

            f=1;

        v.push_back(a);

    }

    if(f)

        return cout<<0<<endl,0;

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

    {

        if(b / 1e18 * v[i] <= 1.00)

            b*=v[i];

        else

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

    }



    if (b <= 1000000000000000000LL)

        cout<<b;

    else

        cout << -1;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.23480557608881x):
#include <stdio.h>



int main(){

 	long long N;

 	scanf("%lld", &N);

 	long long A[N], sum = 1;

 	

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

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

	}

	

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

 		if(A[i] == 0){

 			printf("0\n");

 			return 0;

		 }

	}

	

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

		if(A[i] <= (1000000000000000000/sum)){

			sum *= A[i];

 		} else{

 			printf("-1\n");

            return 0;

		 }

 	}



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

 	

	return 0;

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

using namespace std;



bool overflow(long long a, long long b)

{

    if (a <= (1000000000000000000 / b)) return 0;

    return 1;

}



int main()

{

    long long n;

    cin >> n;

    vector<long long> v(n);

    bool f = 1;

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

        cin >> v[i];

        if (v[i] == 0) f = 0;

    }

    if (f == 0){

        cout << 0 << endl;

        return 0;

    }

    f = 0;

    long long mul = 1LL;

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

        if (overflow(mul, v[i])){

            f = 1;

            break;

        }

        mul *= v[i];

    }

    if (overflow(mul, 1LL)) f = 1;

    if (f) puts("-1");

    else cout << mul << endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.278242680998497x):
#include <stdio.h>



int main(){

 	long long N;

 	scanf("%lld", &N);

 	long long A[N], sum = 1;

 	

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

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

	}

	

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

 		if(A[i] == 0){

 			printf("0\n");

 			return 0;

		 }

	}

	

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

		if(A[i] <= (1000000000000000000/sum)){

			sum *= A[i];

 		} else{

 			printf("-1\n");

            return 0;

		 }

 	}



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

 	

	return 0;

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

using namespace std;

int main()

{

	long long n;

	cin>>n;

	long long a[n];

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

	cin>>a[i];

	long long sum=1,jilu=0;

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

	{

		if(a[i]==0)

		jilu=1;

	}

	if(jilu==1)

	cout<<0;

	else

	{

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

	{

		if((1000000000000000000/a[i])<sum)

		{

			sum=-1;

			break;

		}

	sum*=a[i];

	}

	cout<<sum;

		}

} 

// Faster C++ solution for the same problem (measured speedup: 5.0256749039141555x):
#include <stdio.h>



int main(){

	long long int n;

	scanf("%lld",&n);

	long long int num[n]={};

	long long int res=1;

	long long int temp;

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

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

	}

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

	

		if (num[i]==0){

			printf("0\n");

			return 0;

		}

		}

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

		if(num[i]<=1000000000000000000/res){

			res*=num[i];

		}

		else{

			printf("-1\n");

			return 0;

		}

	}	

		printf("%lld",res);

		return 0;

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

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

#define ll  long long

#define ull unsigned long long

#define IO ios::sync_with_stdio(false), cin.tie(0)

#define endl '\n'

#define main() int main()

#define ff1(i,n) for(int i=1;i<=n;++i)

#define ff0(i,n) for(int i=0;i<n;++i)

#define pi acos(-1.0)

#define ms(a, x) memset(a, x, sizeof(a))

#define lowbit(x) ((x)&(-(x)))

#define pb push_back

#define binarycount __builtin_popcount(x)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define fir first

#define sec second

#define debug(x) cout<<x<<'\n'

#define sa(a) scanf("%d", &a)

#define sab(a, b) scanf("%d%d", &a, &b)

#define pa(a) printf("%d",a)

#define pab(a,b) printf("%d%d",a,b)

#define pii pair<int,int>

using namespace std;



#define maxn 200005

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

const double eps = 1e-8;

ll gcd(ll a, ll b) {return b==0?a:gcd(b, a%b);}

ll lcm(ll a, ll b) {return a/gcd(a, b)*b;}

ll powmod(ll a,ll b){

    ll ans = 1, pow = a%mod;

    while(b) { if(b&1) ans = (ans*pow)%mod; pow = (pow*pow)%mod; b >>= 1; }

    return ans%mod;

}

ull a[maxn];

const ll maxx=1e18;

bool judge(ull a, ull b) {

    if (maxx/a<b) {

        return true;

    }

    return false;

}

main() {

    int n;

    cin>>n;

    int f=1;

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

        cin>>a[i];

        if (a[i]==0) f=0;

    }

    if(f==0){

    	cout<<0<<endl;

    	return 0;

	}

    ll ans=1;

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

        if (judge(ans,a[i])) {

            cout<<"-1"<<endl;

            return 0;

        }

        ans*=a[i];

    }

    cout<<ans<<endl;





}

 

// Faster C++ solution for the same problem (measured speedup: 5.042581767776637x):
#include <stdio.h>



int main(){

	long long int n;

	scanf("%lld",&n);

	long long int num[n]={};

	long long int res=1;

	long long int temp;

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

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

	}

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

	

		if (num[i]==0){

			printf("0\n");

			return 0;

		}

		}

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

		if(num[i]<=1000000000000000000/res){

			res*=num[i];

		}

		else{

			printf("-1\n");

			return 0;

		}

	}	

		printf("%lld",res);

		return 0;

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

#define IO ios::sync_with_stdio(0), cin.tie(0)

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 5;

const ll inf = 1e18;

int main() {

    IO;

    int n;

    ll ans = 1;

    cin >> n;

    bool flag = true;

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

        ll x;

        cin >> x;

        if (x == 0) {

            ans = 0;

            flag = true;

        }

        else if (ULLONG_MAX / ans < x || x * ans > inf) {

            flag = false;

            //cout << ans << " " << ULLONG_MAX << " " << ULLONG_MAX / ans << endl;

        }

        if (flag) ans *= x;

        // cout << ans << endl;

    }

    if (flag)

        cout << ans << endl;

    else {

        cout << -1 << endl;

    }



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.106268177386251x):
#include <stdio.h>



int main(){

	long long int n;

	scanf("%lld",&n);

	long long int num[n]={};

	long long int res=1;

	long long int temp;

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

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

	}

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

	

		if (num[i]==0){

			printf("0\n");

			return 0;

		}

		}

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

		if(num[i]<=1000000000000000000/res){

			res*=num[i];

		}

		else{

			printf("-1\n");

			return 0;

		}

	}	

		printf("%lld",res);

		return 0;

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

using namespace std;

long long int n,a,b,f;

vector<long long int>v;

int main()

{

    cin>>n;

    b=1,f=0;

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

    {

        cin>>a;

        if(a==0)

            f=1;

        v.push_back(a);

    }

    if(f)

        return cout<<0<<endl,0;

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

    {

        if(b / 1e18 * v[i] <= 1.00)

            b*=v[i];

        else

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

    }



    if (b <= 1000000000000000000LL)

        cout<<b;

    else

        cout << -1;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.040399957248737x):
#include <stdio.h>



int main(){

	long long int n;

	scanf("%lld",&n);

	long long int num[n]={};

	long long int res=1;

	long long int temp;

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

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

	}

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

	

		if (num[i]==0){

			printf("0\n");

			return 0;

		}

		}

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

		if(num[i]<=1000000000000000000/res){

			res*=num[i];

		}

		else{

			printf("-1\n");

			return 0;

		}

	}	

		printf("%lld",res);

		return 0;

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

using namespace std;



bool overflow(long long a, long long b)

{

    if (a <= (1000000000000000000 / b)) return 0;

    return 1;

}



int main()

{

    long long n;

    cin >> n;

    vector<long long> v(n);

    bool f = 1;

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

        cin >> v[i];

        if (v[i] == 0) f = 0;

    }

    if (f == 0){

        cout << 0 << endl;

        return 0;

    }

    f = 0;

    long long mul = 1LL;

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

        if (overflow(mul, v[i])){

            f = 1;

            break;

        }

        mul *= v[i];

    }

    if (overflow(mul, 1LL)) f = 1;

    if (f) puts("-1");

    else cout << mul << endl;

}

// Faster C++ solution for the same problem (measured speedup: 5.082223933048308x):
#include <stdio.h>



int main(){

	long long int n;

	scanf("%lld",&n);

	long long int num[n]={};

	long long int res=1;

	long long int temp;

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

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

	}

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

	

		if (num[i]==0){

			printf("0\n");

			return 0;

		}

		}

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

		if(num[i]<=1000000000000000000/res){

			res*=num[i];

		}

		else{

			printf("-1\n");

			return 0;

		}

	}	

		printf("%lld",res);

		return 0;

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

using namespace std;

int main()

{

	long long n;

	cin>>n;

	long long a[n];

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

	cin>>a[i];

	long long sum=1,jilu=0;

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

	{

		if(a[i]==0)

		jilu=1;

	}

	if(jilu==1)

	cout<<0;

	else

	{

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

	{

		if((1000000000000000000/a[i])<sum)

		{

			sum=-1;

			break;

		}

	sum*=a[i];

	}

	cout<<sum;

		}

} 

// Faster C++ solution for the same problem (measured speedup: 5.219056008854531x):
#include <stdio.h>



int main(){

    long long n;

    long long hasil = 1;



    scanf("%lld", &n);

    long long input[n];



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

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

    }

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

        if(input[i] == 0){

            printf("0\n");

            return 0;

        }

    }

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

        if(input[i] <= 1000000000000000000/hasil){

            hasil *= input[i];

        } else {

            printf("-1\n");

            return 0;

        }



    }

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

    return 0;

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

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

#define ll  long long

#define ull unsigned long long

#define IO ios::sync_with_stdio(false), cin.tie(0)

#define endl '\n'

#define main() int main()

#define ff1(i,n) for(int i=1;i<=n;++i)

#define ff0(i,n) for(int i=0;i<n;++i)

#define pi acos(-1.0)

#define ms(a, x) memset(a, x, sizeof(a))

#define lowbit(x) ((x)&(-(x)))

#define pb push_back

#define binarycount __builtin_popcount(x)

#define tr t[root]

#define lson t[root << 1]

#define rson t[root << 1 | 1]

#define fir first

#define sec second

#define debug(x) cout<<x<<'\n'

#define sa(a) scanf("%d", &a)

#define sab(a, b) scanf("%d%d", &a, &b)

#define pa(a) printf("%d",a)

#define pab(a,b) printf("%d%d",a,b)

#define pii pair<int,int>

using namespace std;



#define maxn 200005

const int inf = 0x3f3f3f3f;

const int mod = 1e9+7;

const double eps = 1e-8;

ll gcd(ll a, ll b) {return b==0?a:gcd(b, a%b);}

ll lcm(ll a, ll b) {return a/gcd(a, b)*b;}

ll powmod(ll a,ll b){

    ll ans = 1, pow = a%mod;

    while(b) { if(b&1) ans = (ans*pow)%mod; pow = (pow*pow)%mod; b >>= 1; }

    return ans%mod;

}

ull a[maxn];

const ll maxx=1e18;

bool judge(ull a, ull b) {

    if (maxx/a<b) {

        return true;

    }

    return false;

}

main() {

    int n;

    cin>>n;

    int f=1;

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

        cin>>a[i];

        if (a[i]==0) f=0;

    }

    if(f==0){

    	cout<<0<<endl;

    	return 0;

	}

    ll ans=1;

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

        if (judge(ans,a[i])) {

            cout<<"-1"<<endl;

            return 0;

        }

        ans*=a[i];

    }

    cout<<ans<<endl;





}

 

// Faster C++ solution for the same problem (measured speedup: 5.236613425742688x):
#include <stdio.h>



int main(){

    long long n;

    long long hasil = 1;



    scanf("%lld", &n);

    long long input[n];



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

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

    }

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

        if(input[i] == 0){

            printf("0\n");

            return 0;

        }

    }

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

        if(input[i] <= 1000000000000000000/hasil){

            hasil *= input[i];

        } else {

            printf("-1\n");

            return 0;

        }



    }

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

    return 0;

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

#define IO ios::sync_with_stdio(0), cin.tie(0)

using namespace std;

typedef long long ll;

const int maxn = 1e5 + 5;

const ll inf = 1e18;

int main() {

    IO;

    int n;

    ll ans = 1;

    cin >> n;

    bool flag = true;

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

        ll x;

        cin >> x;

        if (x == 0) {

            ans = 0;

            flag = true;

        }

        else if (ULLONG_MAX / ans < x || x * ans > inf) {

            flag = false;

            //cout << ans << " " << ULLONG_MAX << " " << ULLONG_MAX / ans << endl;

        }

        if (flag) ans *= x;

        // cout << ans << endl;

    }

    if (flag)

        cout << ans << endl;

    else {

        cout << -1 << endl;

    }



    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.302750401394766x):
#include <stdio.h>



int main(){

    long long n;

    long long hasil = 1;



    scanf("%lld", &n);

    long long input[n];



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

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

    }

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

        if(input[i] == 0){

            printf("0\n");

            return 0;

        }

    }

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

        if(input[i] <= 1000000000000000000/hasil){

            hasil *= input[i];

        } else {

            printf("-1\n");

            return 0;

        }



    }

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

    return 0;

}