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"
using namespace std;

typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define REP(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) std::cout << x << '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define sz(x) ((int)(x).size())
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g * b; }
const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };
const int INF = INT_MAX / 2;
const ll  inf = LLONG_MAX / 2;
const int mod = 1000000007;
const int MAX_N = 101010;
const long double PI = acos(-1);

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(10);
	/*------------------------------*/

	ll n;
	cin >> n;
	vector<ll> a(n);
	rep(i, n) cin >> a[i];

	ll g = a[0];
	for (int i = 1; i < n; i++) {
		g = gcd(g, a[i]);
	}

	out(g);
	
	return 0;
}