#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define Inf 2000000001
// a^b
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// a^-1
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

// a^x ≡ b (mod. m) となる最小の正の整数 x を求める
long long modlog(long long a, long long b, int m) {
    a %= m, b %= m;

    // calc sqrt{M}
    long long lo = -1, hi = m;
    while (hi - lo > 1) {
        long long mid = (lo + hi) / 2;
        if (mid * mid >= m) hi = mid;
        else lo = mid;
    }
    long long sqrtM = hi;

    // {a^0, a^1, a^2, ..., a^sqrt(m)} 
    map<long long, long long> apow;
    long long amari = 1;
    for (long long r = 0; r < sqrtM; ++r) {
        if (!apow.count(amari)) apow[amari] = r;
        (amari *= a) %= m;
    }

    // check each A^p
    long long A = modpow(modinv(a, m), sqrtM, m);
    amari = b;
    for (long long q = 0; q < sqrtM; ++q) {
        if (apow.count(amari)) {
            long long res = q * sqrtM + apow[amari];
            if (res > 0) return res;
        }
        (amari *= A) %= m;
    }

    // no solutions
    return -1;
}
	
	
int main(){	

	int t;
	cin>>t;
	
	rep(_,t){
		long long n;
		cin>>n;
		
		cout<<modlog(2,1,2*n-1)<<endl;
	}
	
    return 0;
}