Pythagorean Pair // important

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned ll
#define nl "\n"
#define sz(x) x.size()
#define NumOfDigit(w) log10(w) + 1
#define fill(arr, val)  memset(arr, val , sizeof(arr))
#define PI 3.141592654
#define ceil(w, m) (((w) / (m)) + ((w) % (m) ? 1 : 0))
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define fi first
#define se second
#define cin(v) for (auto&i:v) cin >> i;
#define cout(v) for (auto&i:v) cout << i << " ";
#define fixed(n) fixed << setprecision(n)
#define MOD  1000000007


void IO(){
    ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
}
void files(){
    //freopen("filename.in" , "r" ,stdin);

            #ifndef ONLINE_JUDGE
                     freopen("input.txt", "r", stdin);
                     freopen("output.txt", "w", stdout);
            #endif
}
/*
#include <ext/pb_ds/assoc_container.hpp
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template <typename K, typename V, typename Comp = std::less<K>>
using ordered_map = tree<K, V, Comp, rb_tree_tag, tree_order_statistics_node_update>;
template <typename K, typename Comp = std::less<K>>
using ordered_set = ordered_map<K, null_type, Comp>;

template <typename K, typename V, typename Comp = std::greater_equal<K>>
using ordered_multimap = tree<K, V, Comp, rb_tree_tag, tree_order_statistics_node_update>;
template <typename K, typename Comp = std::greater_equal<K>>
using ordered_multiset = ordered_multimap<K, null_type, Comp>;

*/
bool prime(ll n){
    if(n==2 || n==3) return true;
    if(n<2 || n%2==0  ) return false;
    for(int i=3;i<=sqrt(n);i+=2)   if(n%i==0) return false;
    return true;
}

ll BinEx(ll a , ll b ){
    ll res = 1;
    while (b){
        if (b& 1) res *=a ;
        a*= a;
        b /=2 ;
    }
    return res;
}
void solve(){
ll n , x; cin >> n ;
x=n;
int cnt2=0;
while(n%2==0){
    cnt2++;  // get how many times n is divisble by 2
    n/=2 ;
}
// odd /
if(cnt2&1) n*=2 , cnt2--; // to be perfect
// 2^ ( even ) -> perfect square
cnt2/=2 ;   // as we * twice taht
ll ans=BinEx(2,cnt2);

for(ll i=0;i*i <=n; i++){ // n become smaller element after /2
    ll tmp = n - (i*i);
    ll sq=sqrt(tmp);
     // perfect square * perfect square = perfect square
    if(sq*sq + i*i == n )
        return void (cout << i * ans  << " " <<
        sq * ans  <<"\n" );

}

cout << -1 << nl;

}
int main(){
                 IO();          files();

    int testCase=1;  // one test case
       cin >> testCase ;
 while(testCase--)
        solve();  // my code

    return 0;
}