#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return true; } return false; }
#define vi vector<int>
#define vl vector<ll>
#define vii vector<pair<int,int>>
#define vll vector<pair<ll,ll>>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>>
#define vvii vector<vector<pair<int,int>>>
#define vvll vector<vector<pair<ll,ll>>>
#define vst vector<string>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define mkunique(x) sort(all(x));(x).erase(unique(all(x)),(x).end())
#define fi first
#define se second
#define mp make_pair
#define si(x) int(x.size())
const int mod=998244353,MAX=10000005,INF=15<<26;

//高速素因数分解

/**
 * Author: chilli, Ramchandra Apte, Noam527, Simon Lindholm
 * Date: 2019-04-24
 * License: CC0
 * Source: https://github.com/RamchandraApte/OmniTemplate/blob/master/modulo.hpp…
 * Description: Calculate $a\cdot b\bmod c$ (or $a^b \bmod c$) for $0 \le a, b \le c \le 7.2\cdot 10^{18}$.
 * Time: O(1) for \texttt{modmul}, O(\log b) for \texttt{modpow}
 * Status: stress-tested, proven correct
 * Details:
 * This runs ~2x faster than the naive (__int128_t)a * b % M.
 * A proof of correctness is in doc/modmul-proof.tex. An earlier version of the proof,
 * from when the code used a * b / (long double)M, is in doc/modmul-proof.md.
 * The proof assumes that long doubles are implemented as x87 80-bit floats; if they
 * are 64-bit, as on e.g. MSVC, the implementation is only valid for
 * $0 \le a, b \le c < 2^{52} \approx 4.5 \cdot 10^{15}$.
 */
#pragma once

typedef unsigned long long ull;
ull modmul(ull a, ull b, ull M) {
    ll ret = a * b - M * ull(1.L / M * a * b);
    return ret + M * (ret < 0) - M * (ret >= (ll)M);
}
ull modpow(ull b, ull e, ull mod) {
    ull ans = 1;
    for (; e; b = modmul(b, b, mod), e /= 2)
        if (e & 1) ans = modmul(ans, b, mod);
    return ans;
}

/**
 * Author: chilli, SJTU, pajenegod
 * Date: 2020-03-04
 * License: CC0
 * Source: own
 * Description: Pollard-rho randomized factorization algorithm. Returns prime
 * factors of a number, in arbitrary order (e.g. 2299 -> \{11, 19, 11\}).
 * Time: $O(n^{1/4})$, less for numbers with small factors.
 * Status: stress-tested
 *
 * Details: This implementation uses the improvement described here
 * (https://en.wikipedia.org/wiki/Pollard%27s_rho_algorithm#Variants…), where
 * one can accumulate gcd calls by some factor (40 chosen here through
 * exhaustive testing). This improves performance by approximately 6-10x
 * depending on the inputs and speed of gcd. Benchmark found here:
 * (https://ideone.com/nGGD9T)
 *
 * GCD can be improved by a factor of 1.75x using Binary GCD
 * (https://lemire.me/blog/2013/12/26/fastest-way-to-compute-the-greatest-common-divisor/…).
 * However, with the gcd accumulation the bottleneck moves from the gcd calls
 * to the modmul. As GCD only constitutes ~12% of runtime, speeding it up
 * doesn't matter so much.
 *
 * This code can probably be sped up by using a faster mod mul - potentially
 * montgomery reduction on 128 bit integers.
 * Alternatively, one can use a quadratic sieve for an asymptotic improvement,
 * which starts being faster in practice around 1e13.
 *
 * Brent's cycle finding algorithm was tested, but doesn't reduce modmul calls
 * significantly.
 *
 * Subtle implementation notes:
 * - we operate on residues in [1, n]; modmul can be proven to work for those
 * - prd starts off as 2 to handle the case n = 4; it's harmless for other n
 *   since we're guaranteed that n > 2. (Pollard rho has problems with prime
 *   powers in general, but all larger ones happen to work.)
 * - t starts off as 30 to make the first gcd check come earlier, as an
 *   optimization for small numbers.
 */
#pragma once

/**
 * Author: chilli, c1729, Simon Lindholm
 * Date: 2019-03-28
 * License: CC0
 * Source: Wikipedia, https://miller-rabin.appspot.com
 * Description: Deterministic Miller-Rabin primality test.
 * Guaranteed to work for numbers up to $7 \cdot 10^{18}$; for larger numbers, use Python and extend A randomly.
 * Time: 7 times the complexity of $a^b \mod c$.
 * Status: Stress-tested
 */
#pragma once

bool isPrime(ull n) {
    if (n < 2 || n % 6 % 4 != 1) return (n | 1) == 3;
    ull A[] = {2, 325, 9375, 28178, 450775, 9780504, 1795265022},
    s = __builtin_ctzll(n-1), d = n >> s;
    for (ull a : A) {   // ^ count trailing zeroes
        ull p = modpow(a%n, d, n), i = s;
        while (p != 1 && p != n - 1 && a % n && i--)
            p = modmul(p, p, n);
        if (p != n-1 && i != s) return 0;
    }
    return 1;
}

ull pollard(ull n) {
    auto f = [n](ull x) { return modmul(x, x, n) + 1; };
    ull x = 0, y = 0, t = 30, prd = 2, i = 1, q;
    while (t++ % 40 || __gcd(prd, n) == 1) {
        if (x == y) x = ++i, y = f(x);
        if ((q = modmul(prd, max(x,y) - min(x,y), n))) prd = q;
        x = f(x), y = f(f(y));
    }
    return __gcd(prd, n);
}
vector<ull> factor(ull n) {
    if (n == 1) return {};
    if (isPrime(n)) return {n};
    ull x = pollard(n);
    auto l = factor(x), r = factor(n / x);
    l.insert(l.end(), all(r));
    return l;
}

vector<int> prime;//i番目の素数
bool is_prime[MAX+1];

void sieve(int n){
    for(int i=0;i<=n;i++){
        is_prime[i]=true;
    }
    
    is_prime[0]=is_prime[1]=false;
    
    for(int i=2;i<=n;i++){
        if(is_prime[i]){
            prime.push_back(i);
            for(int j=2*i;j<=n;j+=i){
                is_prime[j] = false;
            }
        }
    }
}


int main(){
    
    std::ifstream in("text.txt");
    std::cin.rdbuf(in.rdbuf());
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    sieve(MAX-2);
    vl X;
    for(ll p:prime){
        if(is_prime[p+2]){
            X.pb(p*(p+2));
        }
    }
    
    int Q;cin>>Q;
    while(Q--){
        ll N;cin>>N;
        auto it=upper_bound(all(X),N);
        if(it!=X.begin()){
            it--;
            cout<<(*it)<<"\n";
        }else{
            cout<<-1<<"\n";
        }
    }
}