#pragma GCC optimize("O3") #include #define ll long long #define rep(i,n) for(ll i=0;i<(n);i++) #define pll pair #define pii pair #define pq priority_queue #define pb push_back #define eb emplace_back #define fi first #define se second #define endl '\n' #define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0); #define lb(c,x) distance(c.begin(),lower_bound(all(c),x)) #define ub(c,x) distance(c.begin(),upper_bound(all(c),x)) using namespace std; const ll INF=1e18; ll mod; struct mint { ll x; // typedef long long ll; mint(ll x=0):x((x%mod+mod)%mod){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;} mint operator+(const mint a) const { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; istream& operator>>(istream& is, mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} // combination mod prime // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619 int main(){ ll n; cin >> n; rep(i,n){ ll p; cin >> p; mod=p; ll x=2; mint val=2; ll ans=-1; while(x<=INF){ mint s=val.pow(x); if(s.x==(x%mod)){ ans=x; break; } x*=2; } cout << ans << endl; } return 0; }