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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
#define ll                  long long int
#define ull                 unsigned long long int
#define ld                  long double
#define fi                  first
#define se                  second
#define pb                  push_back
#define pbb                 pop_back
#define mp                  make_pair
#define popcount            __builtin_popcountll
#define pii     	        pair<int,int>
#define all(x)              x.begin(), x.end()
#define uniq(x)             x.erase(unique(x.begin(),x.end()),x.end())
#define mem(array, value)   memset(array, value, sizeof(array))
#define lcm(a,b)            (abs(a)/gcd(a,b))*abs(b)
#define random(a, b)        ((((rand() << 15) ^ rand()) % ((b) - (a) + 1)) + (a))
#define PI      	        2*acos(0.0)
#define INF     	        1e9
#define EPS     	        1e-8
#define line                cout << "\n==========\n"
#define fastRead 	        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> orderedSet;

//int dx[] = {1, -1, 0, 0},                  dy[] = {0, 0, 1, -1};                  // 4 Direction
//int dx[] = {1, -1, 0, 0, 1, 1, -1, -1},    dy[] = {0, 0, 1, -1, 1, -1, 1, -1};    // 8 Direction
//int dx[] = {1, -1, 1, -1, 2, 2, -2, -2},   dy[] = {2, 2, -2, -2, 1, -1, 1, -1};   // Knight Direction

inline ll gcd(ll a, ll b){
    a = abs(a);
    b = abs(b);
    while (b){
        a = a % b;
        swap (a, b);
    }
    return a;
}
inline ll power(ll a, ll p){
    ll res = 1, x = a;
    while (p){
        if (p & 1){
            res = (res * x);
        }
        x = (x * x);
        p >>= 1;
    }
    return res;
}
inline ll bigmod(ll a, ll p, ll m){
    ll res = 1 % m, x = a % m;
    while (p){
        if (p & 1){
            res = (res * x) % m;
        }
        x = (x * x) % m;
        p >>= 1;
    }
    return res;
}

int main ()
{
    #ifdef Lollipop
    //freopen ("input.txt", "r", stdin);
    //freopen ("output.txt", "w", stdout);
    #endif
    int n;
    cin >> n;
    int ans = 0;
    int x;
    for (int i = 0; i<n; i++){
        cin >> x;
        ans = gcd(ans, x);
    }
    cout << ans << endl;
    return 0;
}