Untitled

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define nl "\n"
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define pb push_back
#define llmx LONG_LONG_MAX
#define llmn LONG_LONG_MIN
#define mod 1000000007
const int N = 1e5 + 7;
const int INF = 1e9 + 10;

bool predicate(double mid, ll x1, ll y1, ll x2, ll y2, ll ww, ll hh, ll w, ll h) {
    y1 -= mid;
    y2 -= mid;
    ll rem_up = h - y2; 
    ll rem_down = y1 - 0;
    if (rem_up >= hh || rem_down >= hh) return true;

    y1 += mid * 2;  
    y2 += mid * 2;
    ll rem_up2 = h - y2; 
    ll rem_down2 = y1 - 0;
    if (rem_up2 >= hh || rem_down2 >= hh) return true;

    y1 -= mid;  
    y2 -= mid;
    x1 -= mid;
    x2 -= mid;
    ll rem_right = w - x2;
    ll rem_left = x1 - 0;
    if (rem_right >= ww || rem_left >= ww) return true;

    x1 += mid * 2;  
    x2 += mid * 2;
    ll rem_right2 = w - x2;
    ll rem_left2 = x1 - 0;
    if (rem_right2 >= ww || rem_left2 >= ww) return true;

    return false;
}


void solve(){
    ll w, h;
    cin >> w >> h;
    ll x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    ll ww, hh;
    cin >> ww >> hh;

    double l = 0;
    double r = 1e8;
    double ans = -1;
    while (r - l > 1e-6) {
        double mid = l + (r - l) / 2;
        if (predicate(mid, x1, y1, x2, y2, ww, hh, w, h)) {
            ans = mid;
            r = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }    
    cout << ans << nl;
}

/* Hey you should check this out
    * int overflow, array bounds
    * reset global array and variable
    * look for special cases (n=1?)
    * do something instead of nothing and stay organized
    * bruteforce to find pattern
    * DON'T GET STUCK ON ONE APPROACH
    * Think the problem backwards
    * In practice time don't see failing test case
*/

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t--) {
        solve();
    }
    return 0;
}