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)

#include <bits/stdc++.h>
using namespace std;

const int64_t MOD = 1000000007;
typedef long long ll;
typedef unsigned long long ull;

#define FOR(i, start, end) for(uint64_t i=start; i<end; i++)
#define REP(i, n) FOR(i, 0, n)

// 最大公約数gcd
// 最小公倍数lcm=m*n/gcd
uint64_t gcd(uint64_t m, uint64_t n) {
  uint64_t temp;
  while (m % n != 0){
    temp = n;
    n = m % n;
    m = temp;
  }
  return n;
}

uint64_t lcm(uint64_t m, uint64_t n) {
  return (m*n)/gcd(m,n);
}

// vector<vector<uint64_t> > v(n+1, vector<uint64_t>(n+1, 0))
// v[n][k]に組み合わせ数が入る。
void comb(vector<vector <uint64_t> > &v){
  for(uint64_t i = 0;i <v.size(); i++){
    v[i][0]=1;
    v[i][i]=1;
  }
  for(uint64_t k = 1;k <v.size();k++){
    for(uint64_t j = 1;j<k;j++){
      v[k][j]=(v[k-1][j-1]+v[k-1][j]);
    }
  }
}

// 掛け算オーバーフロー判定
bool is_product_overflow(uint64_t a, uint64_t b) {

	uint64_t prod = a * b;
	
	return (prod / b != a);
}

//素因数分解
void primeFactorization(uint64_t a, list<uint64_t> &factors){ //素因数分解を出力するプログラム
  long i,sq;
  if(a%2==0){ //偶数の場合
    factors.push_back(2);
    primeFactorization(a/2,factors); //2で割った値で再帰
    return;
  }
  sq = sqrt(a);
  for(i=3;i<=sq;i+=2){ //3以上√a以下の奇数の場合
	if(a%i==0){
	  factors.push_back(i);
      primeFactorization(a/i,factors); //割れた値で再帰
      return;
	}
  }
  //偶数でも3以上√a以下の奇数の場合でも割り切れない場合
  if(a!=1){ //aが1でないなら、a自身は素数
    factors.push_back(a);
  }
}

// フェルマーの小定理
// mod. m での a の逆元 a^{-1} を計算する
// (a/b)%m = a*movinv(b,m)%m  
int64_t modinv(int64_t a, int64_t m) {
    int64_t b = m, u = 1, v = 0;
    while (b) {
        int64_t t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

// 円周率
// M_PI

// #include <iomanip> // setprecisionを使用するのに必要
// cout << std::fixed << std::setprecision(15) << y << endl;

// 昇順
// priority_queue<int, vector<int>, greater<int> > queue;

signed main() {
  int n;
  cin >> n;
  
  ull a;
  cin >> a;
  ull gc = a;
  for(int i=1;i<n;i++){
    cin >> a;
    gc = gcd(a,gc);
  }
  cout << gc << endl;

  return 0;
}