#include<bits/stdc++.h>
using namespace std;
int a[100];
int pos = -1;

void binarysearch(int l, int r, int x){
	if(l > r)return;
	else{
		int mid = (l+r)/2;
		if(x == a[mid]) {
			pos = mid;
			return;
		}
		if(x < a[mid]) return binarysearch(l, mid-1, x);
		else return binarysearch(mid+1, r, x);
	}
}

int main(){
	int n;
	cin>>n; //輸入元素個數(shù) 
	for(int i=0; i<n; i++)cin>>a[i];
	sort(a, a+n);
	binarysearch(0, n-1, 5);//二分搜索 
	cout<<pos;//輸出找到的位置下標(biāo) 
	return 0;
} 

#include<bits/stdc++.h>
using namespace std;
long Ackerman(long n, long m){
	if(n >= 0 && m >= 0){
		if(n == 1 && m == 0)return 2;
		if(n == 0 && m >= 0)return 1;
		if(n >= 2 && m == 0)return n+2;
		if(n >= 1 && m >= 1)return Ackerman(Ackerman(n-1, m), m-1);
	}
} 

int main(){
	long n, m;
	cin>>n>>m;
	cout<<Ackerman(n, m);
	return 0;
} 

#include<bits/stdc++.h>
using namespace std;
int fibonacci(int n){
	if(n == 1)return 1;
	else if(n == 2)return 1;
	else return fibonacci(n-1)+fibonacci(n-2);
	
}
int main(){
	int n;
	cin>>n;//返回斐波那契數(shù)列第幾項 
	cout<<fibonacci(n);
	return 0;
} 

#include<bits/stdc++.h>
using namespace std;
int a[20], b[20];//a[]為排列的盒子，b[]為判斷元素是否放過的數(shù)組 
int n;
void perm(int k){//k表示開始放第k個數(shù) 
	if(k == n+1){//當(dāng)k>n時說明第k個數(shù)已經(jīng)放好，已經(jīng)一組排列完畢 
		for(int i=1; i<=n; i++){
			cout<<a[i]<<" ";
		}
		cout<<endl;
	}
	else{
		for(int i=1; i<=n; i++){//1-n個數(shù)各自放入盒子a[]中 
			if(b[i] == 0){//初始化b[]都為0，為0說明這個數(shù)沒放過 
				a[k] = i;//第k個數(shù)放入i 
				b[i] = 1;//i放了因此后面不能再放了 
				perm(k+1);//放第二個數(shù)。 
				b[i] = 0;//雖然第一次排列放了，但是第二次排列還需要用到。 
			}
		}
	}
}
int main(){
	cin>>n;//排列數(shù)個數(shù) 
	perm(1);//從放第一個數(shù)開始排列 
	return 0;
} 

#include<bits/stdc++.h>
using namespace std;
int gcd(int a, int b){
	return b==0?a:gcd(b, b%a);
} 
int main(){
	int a, b;
	cin>>a>>b;
	cout<<gcd(a,b);
} 

#include<bits/stdc++.h>
using namespace std;
long mul(long x, long y, long n){
	if(x == 0 || y == 0)return 0;
	else if(n == 1)return x * y;
	else{
		long A = (long)x / pow(10, (long)(n/2));
		long B = x - A * pow(10, n/2);
		long C = (long)y / pow(10, (long)(n/2));
		long D = y - C * pow(10, n/2);
		long AC = mul(A, C, n/2);
		long BD = mul(B, D, n/2);
		long A_BD_C = mul((A - B),(D - C), n/2); 
		return AC * pow(10, n) + (A_BD_C + AC + BD)* pow(10, (long)(n/2)) + BD;
	}
}
int main(){
	long a, b, n, sign;
	if((a<0 && b>0) || (a>0 && b<0))sign = -1;
	else sign = 1;
	cin>>a>>b>>n;
	cout<<mul(a, b, n) * sign;
}