Problem:
Score : 300 points

Problem Statement

There are N monsters, numbered 1, 2, ..., N.

Initially, the health of Monster i is A_i.

Below, a monster with at least 1 health is called alive.

Until there is only one alive monster, the following is repeated:

A random alive monster attacks another random alive monster.

As a result, the health of the monster attacked is reduced by the amount equal to the current health of the monster attacking.

Find the minimum possible final health of the last monster alive.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible final health of the last monster alive.

Sample Input 1

4
2 10 8 40

Sample Output 1

2

When only the first monster keeps on attacking, the final health of the last monster will be 2, which is minimum.

Sample Input 2

4
5 13 8 1000000000

Sample Output 2

1

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

1000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

				/*****************************************************
 				*	Logged in : HABIBUR RAHMAN						 *
				*	Dept	  : COMPUTER SCIENCE AND ENGINEERING	 *
				*													 *
				*	KHULNA UNIVERSITY OF ENGINEERING AND TECHNOLOGY  *
				******************************************************/


#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef vector<string> vs;
typedef pair<ll,ll> pll;
#define fast_io 			ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define all(v) 				v.begin(),v.end()
#define allr(v) 			v.rbegin(),v.rend()
#define rep(i,a,b) 			for(ll i=a;i<=b;i++)
#define rep0(i,n)			for(ll i=0;i<n;i++)
#define rrep0(i, n) 		for(ll i=(ll)(n);i>0;--i)
#define rrep(i, a, b) 		for(ll i=(ll)(a);i>=(ll)(b);--i)
#define pb(z) 				emplace_back(z)
#define ff					first
#define ss 					second
#define caset 				int t;cin>>t;for(int tt=1;tt<=t;tt++)
#define sz(X) 				((ll)(X).size())
#define scan(X) 			scanf("%s", (X))
#define ms(v,val)			memset((v),(val),sizeof((v)))
#define ms0(X) 				memset((X), 0, sizeof((X)))
#define ms1(X) 				memset((X), -1, sizeof((X)))
#define m_p(a,b)            make_pair(a,b)
#define get_pos(c,x) 		(lower_bound(c.begin(),c.end(),x)-c.begin())
#define lcm(a,b)            (a*b)/__gcd(a,b)
#define INF9                1000000010
#define INF18               4000000000000000010
#define bpll(a)             __builtin_popcountll(a)
#define sqr(a)              ((a)*(a))
#define nl					"\n"
//#define mod					1e9+7;
#define mx_el(v)			*max_element(all(v));
#define mn_el(v)			*min_element(all(v));
#define pcase(z,x)    		cout<<"Case "<<z<<": "<<x<<"\n"
const double pi = 2*acos(0) ;
#define display(v)			vi :: iterator it;for(it=(v).begin();it!=(v).end();it++){cout << *it << " " ;}cout << endl;
#define taking(n,v) 		for(ll i=0;i<(n);i++){ll inp;cin>>inp;(v).push_back(inp);}
/*-------------------------some important functions--------------------------------------------------------------------*/
ull binpow(ull b,ull n){if(n==0)return 1;if(n==1)return b;if(n&1) return binpow(b,n-1)*b;else return binpow(b*b,n/2);}
ll bigmod(ll a,ll n,ll mod){if(n==0)return 1;ll res=bigmod(a,n/2,mod);if(n&1){return ((res%mod)*(res%mod)*a)%mod;}else return ((res%mod)*(res%mod))%mod;}
/*--------------------------------------------x----------------------------------------------x--------------------------*/
void start(){
      #ifndef ONLINE_JUDGE 
      freopen("input.txt" , "r" , stdin) ;
      freopen("output.txt" , "w" , stdout) ;
      #endif
}
void solve(){
	ll n;cin>>n;
	ll a[n+1];
	rep0(i,n){
		cin>>a[i];
	}
	ll cnt=a[0];
	rep(i,1,n-1){
		cnt=__gcd(a[i],cnt);
	}
	cout << cnt << nl;
}
 	 
int main() {
	fast_io
	//start();
    //caset{
    	//cout << "Case " << tt << ": ";
		  solve();
	//}
	
	return 0;
}