//#define //_GLIBCXX_DEBUG
#include <bits/stdc++.h>
#include <deque>
#include <atcoder/all>
#pragma GCC optimize("03")
using namespace std;
using namespace atcoder;
using mint = atcoder::modint1000000007;
using ll = long long;
using ull = unsigned long long;
using vll=vector<ll>;
using vvll=vector<vector<ll>>;
using vvvll=vector<vvll>;
using vvvvll=vector<vvvll>;
using Graph=vvll;
using Edgegraph=vector<vector<pair<ll,ll>>>;
using vch=vector<char>;
using vvch=vector<vector<char>>;
using P=pair<ll,ll>;
using vP=vector<P>;
using tup=tuple<ll,ll,ll>;
using vbl=vector<bool>;
using vvbl=vector<vbl>;
using vs=vector<string>;
using vvs=vector<vs>;
using vd=vector<double>;
using vvd=vector<vd>;
using vvld=vector<vector<long double>>;
const int infint = 1073741823;
const ll inf = 1LL << 60;
template <class T> inline bool chmax(T& a,T b){if (a<b){a=b;return 1;}return 0;}
template <class T> inline bool chmin(T& a,T b){if (a>b){a=b;return 1;}return 0;}
#define rep(i,x,lim) for(ll i = (x);i < (ll)(lim);i++)
#define rep2(j,x,lim) for(int j = (x);j < (int)(lim);j++)
const ll big=(1e+9)+7;
const ll big2=998244353;

ll dx[8]={1,-1,0,0,1,1,-1,-1};
ll dy[8]={0,0,1,-1,1,-1,1,-1};

int modpow(ll x,ll n,ll m){
    if(n==0) return 1%m;
    x=((x%m)+m)%m;
    if(n%2==0){
        ll r=modpow(x,n/2,m);
        return r*r%m;
    }
    else{
        ll r=modpow(x,n/2,m);
        return r*r%m*x%m;
    }
}
//pは素数でなければならない。
int revmod(ll x,ll p){return modpow(x,p-2,p);}
//99 のRを高速に求める
int modp(ll p,ll q){
    ll gc=gcd(p,q);
    p/=gc;q/=gc;
    ll rev=revmod(p,big2);
    return (rev*q)%big2;
}
//nCrを求める modbig2
int nCr(ll n,ll r){
    ll ans=1;
    rep(i,1,n+1) ans=(ans*i)%big2;
    rep(i,1,r+1) ans=(ans*modpow(i,big2-2,big2))%big2;
    rep(i,1,n-r+1) ans=(ans*modpow(i,big2-2,big2))%big2;
    return ans;
}

#include <iostream>
using namespace std;

// AtCoder の modint を使う

const int MAX = 510000;
mint fac[MAX], finv[MAX], inv[MAX];

void COMinit() {
    const int MOD = mint::mod();
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i;
        inv[i] = MOD - inv[MOD%i] * (MOD / i);
        finv[i] = finv[i - 1] * inv[i];
    }
}

// 二項係数計算
mint COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * finv[k] * finv[n - k];
}

inline long long mod(long long a, long long m) {
    return (a % m + m) % m;
}

long long extGcd(long long a, long long b, long long &p, long long &q) {  
    if (b == 0) { p = 1; q = 0; return a; }  
    long long d = extGcd(b, a%b, q, p);  
    q -= a/b * p;  
    return d;  
}

// 中国剰余定理
// リターン値を (r, m) とすると解は x ≡ r (mod. m)
// 解なしの場合は (0, -1) をリターン
pair<long long, long long> ChineseRem(const vector<long long> &b, const vector<long long> &m) {
  long long r = 0, M = 1;
  for (int i = 0; i < (int)b.size(); ++i) {
	long long p, q;
	long long d = extGcd(M, m[i], p, q); // p is inv of M/d (mod. m[i]/d)
	if ((b[i] - r) % d != 0) return make_pair(0, -1);
	long long tmp = (b[i] - r) / d * p % (m[i]/d);
	r += M * tmp;
	M *= m[i]/d;
  }
  return make_pair(mod(r, M), M);
}

// 入力：n <= 4.5*10^18, nは平方数
int64_t isqrt(int64_t n)
{
    // 4.5*10^18 < 2^62
    // 真の答えは 2 <= _ < 2^31 の範囲にある
    int64_t low = 0, high = 1LL<<30;
    while (low < high-1) {
        int64_t mid = (low + high) / 2;
        int64_t mid2 = mid * mid; // 最初の上界を2^31程度に抑えているのでここではオーバーフローしない
        if (mid2 < n) {
            low = mid;
        } else if (mid2 == n) {
            return mid;
        } else {
            high = mid;
        }
    }
    return low;
}

using S=array<mint,3>;
S e(){return S();}
S op(S x,S y){return {x[0]+y[0],x[1]+y[1],x[2]+y[2]};}

int main(){
    ll T;
    cin >> T;
    while(T--){
        ll N;
        cin >> N;
        ll first=inf;
        ll second=inf;
        for(ll i=0;i<=29;i++){
            if(N >> i & 1){
                if(first==inf) first=(1 << i);
                else if(second==inf) second=(1<< i);
            }
        }
        if(second!=inf) cout << second-first << '\n';
        else cout << -1 << '\n';
    }
}