Problem:
Score : 200 points

Problem Statement

You are given positive integers A and B.

Find the K-th largest positive integer that divides both A and B.

The input guarantees that there exists such a number.

Constraints

All values in input are integers.

1 \leq A, B \leq 100

The K-th largest positive integer that divides both A and B exists.

K \geq 1

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the K-th largest positive integer that divides both A and B.

Sample Input 1

8 12 2

Sample Output 1

2

Three positive integers divides both 8 and 12: 1, 2 and 4.
Among them, the second largest is 2.

Sample Input 2

100 50 4

Sample Output 2

5

Sample Input 3

1 1 1

Sample Output 3

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define ll long long int
#define vi vector<int>
#define vl vector<ll>
#define v2i vector<vector<int>>
#define v2l vector<vector<ll>>
#define ppi pair<int,int>
#define vpi vector<ppi>
#define ppl pair<ll,ll>
#define all(x) x.begin(),x.end()
#define ii int,int 
#define ff first    
#define ss second
#define forn(i,a,b) for(int i=a;i<b;i++)
#define forr(i,a,b) for(int i=a;i>=b;i--)
#define forv(i,m) for(auto i : m)
#define imx INT_MAX
#define imn INT_MIN
#define inf LONG_MAX
#define minf LONG_MIN
#define endl "\n"
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);
#define sze size()
#define rvs reverse 
#define itr iterator 
#define pre cout<<fixed<<setprecision(10);
#define umap unordered_map
#define uset unordered_set

/*
#define MODM(a,b,md) ((a%md)*(b%md))%md
#define MODA(a,b,md) ((a%md)+(b%md))%md
#define ull unsigned long long int
#define si short int
*/

/*
bool compare(int a,int b) {return a > b;}
bool compare1(ppi a,ppi b) {return a.ff > b.ff;}
bool compare2(ppi a,ppi b) {return a.ss < b.ss;}
ll Npower(ll a,ll b) {ll product = 1; while(b){if(b&1){product = product*a;}a = a*a;b = b>>1;} return product;}
ll power(ll a,ll b,ll md) {ll product = 1; while(b){if(b&1){product = MODM(product,a,md);}a = MODM(a,a,md);b = b>>1;} return product%md;}
ll GCD(ll a,ll b) {return b==0 ? a:GCD(b,a%b);}
ll LCM(ll a,ll b) {return (a/GCD(a,b))*b; }
*/

int main()
{
    fast


        
    int a,b,c;
    cin>>a>>b>>c;


    int gcd = __gcd(a,b);

    vi v;

    int i = 1;
    for(; i*i < gcd; i++)
    {
        if(gcd%i == 0)
        {
            v.pb(i);
            v.pb(gcd/i);
        }
    }

    if(i*i == gcd) v.pb(i);

    sort(all(v));


    cout<<v[c-1];




    return 0;
}