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/*
===========================================================
* @Name: 1043F Make It One
* @Author: Shawn
* @create Time: 2020/3/8 11:49:59
* @url: https://codeforces.com/contest/1043/problem/F
* @Description: 容斥原理，组合数比较大，Pascal公式超内存，需要逆元计算组合数
===========================================================
*/
#ifdef ONLINE_JUDGE
#define NDEBUG //ban assert when submit to online judge
#endif
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxA = 3e5 + 1;
const ll MOD = 1e9 + 7;
//计算组合数：从a个数中取b个处理有多少种方案（Pascal公式，帕斯卡恒等式）
class CombinationPascal
{
private:
vector<vector<ll>> c;
/**
* 从M个数中取N个处理有多少种方案,动态规划（Pascal公式，帕斯卡恒等式）
* https://zhuanlan.zhihu.com/p/74787475
*/
void InitData(ll M)
{
c[0][0] = 1;
for (ll i = 0; i <= M; ++i)
{
for (ll j = 0; j <= i; ++j)
{
if (j == 0)
c[i][j] = 1;
else if (i == 1)
c[i][j] = 1;
else if (i == j)
c[i][j] = 1;
else
c[i][j] = ((c[i - 1][j - 1] % MOD) + (c[i - 1][j] % MOD) % MOD) % MOD;
}
}
}
public:
CombinationPascal(int m)
{
c.assign(m + 1, vector<ll>(m + 1, 0));
InitData(m);
}
/**
* 从a个数中取b个处理有多少种方案（Pascal公式，帕斯卡恒等式）
*/
ll C(ll a, ll b)
{
if (b > a)
return 0;
else if (b == 0)
return 1;
else if (a == 1)
return 1;
else if (a == b)
return 1;
else
return c[a][b];
}
};
//计算组合数：从M个数中取N个处理有多少种方案（逆元,Modular Multiplicative Inverse）
//https://www.zybuluo.com/ArrowLLL/note/713749
class CombinationMMI
{
private:
vector<ll> factorial;
void calFactorial() //求maxA以内的数的阶乘
{
factorial[0] = 1;
factorial[1] = 1;
for(ll i = 2; i < maxA+1; ++i)
factorial[i] = factorial[i - 1] * i % MOD;
}
//费马小定理求逆元
ll pow(ll a, ll n, ll p) //快速幂 a^n % p
{
ll ans = 1;
while(n)
{
if(n & 1) ans = ans * a % p;
a = a * a % p;
n >>= 1;
}
return ans;
}
ll mmi(ll a, ll b) //费马小定理求逆元
{
return pow(a, b - 2, b);
}
public:
CombinationMMI()
{
factorial.assign(maxA+1,0);
calFactorial();
}
ll C(ll a, ll b) //计算C(a, b)
{
if(b>a)
return 0;
return (factorial[a] * mmi(factorial[b], MOD) % MOD * mmi(factorial[a - b], MOD) % MOD) % MOD;
}
};
void solve()
{
int n;
cin >> n;
//assert(n > 0);
CombinationMMI CMMI;
//CombinationPascal CPAS(n);
vector<int> a(n, 0);
int gcd = 0;
int mx = 0;
for (int i = 0; i <= n - 1; ++i)
{
cin >> a[i];
//++m[a[i]];
//++f[1][a[i]];
gcd = __gcd(gcd, a[i]);
mx = max(mx, a[i]);
}
//vector<vector<int>> f(7 + 1, vector<int>(mx + 1, 0));
vector<int> f2(mx + 1, 0);
vector<int> m(mx, 0);
for (int i = 0; i <= n-1 ; ++i)
++m[a[i]];
for (int i = 1; i <= mx; ++i)
for (int j = i + i; j <= mx; j += i)
m[i] += m[j];
if (gcd > 1)
{
cout << -1 << '\n';
return;
}
for (int i = 1; i <= 7; ++i)
{
for (int j = mx; j >= 1; --j)
{
f2[j] = 1ll * CMMI.C(m[j],i) % MOD;
for (int k = j + j; k <= mx; k += j)
{
f2[j] -= f2[k];
if (f2[j] < 0)
f2[j] += MOD;
}
}
if (f2[1] > 0)
{
cout << i << '\n';
return;
}
}
}
int main()
{
ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("CF_1043F_MakeItOne.in", "r", stdin);
//freopen("CF_1043F_MakeItOne.out", "w", stdout);
#endif
solve();
// Test
// int n=100;
// CombinationMMI CMMI;
// CombinationPascal CPAS(n);
// for (int i = 0; i <= n - 1; ++i)
// for (int j = 0; j <= i; ++j)
// cout << i << ":" <<setw(3)<< j << " " << CMMI.C(i,j) << " " <<setw(10)<< CPAS.C(i,j) << "\n";
cout.flush();
return 0;
}
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