游戏钦点 构造

题号：NC21705
游戏钦点

构造题

1，计算出总比赛次数 k
2，特判 x == 2 || y == 2 情况，一旦出现，必不可能，不能只判断x，还要判断y 坑！
3，贪心求最小次数，从大往下枚举，能用就用，可递归（特判 2 这个数），也可从大到小，不能处理时 该位置必定大于 剩余的x
只需 判断一下x的奇偶即可。

/*

hello world!

Just do it!

start time:2022-04-28 10:43:15.393571

*/

// #pragma GCC optimize (2)
// #pragma G++ optimize (2)
#include 
#define zero(x) memset(x, 0, sizeof(x));
#define one(x) memset(x, -1, sizeof(x));
#define m_INF(x) memset(x, 0x3f, sizeof(x));
#define m_inf(x) memset(x, 0x3f, sizeof(x));
#define m_f_INF(x) memset(x, -0x3f, sizeof(x));
#define m_f_inf(x) memset(x, -0x3f, sizeof(x));
#define all(x) x.begin(), x.end()
#define endl "\n" 
#define fi first
#define se second
#define lbt(x) ((x)&(-x))
#define pb push_back

struct cmpl{ template  bool operator()(const A &a1, const B &b1) { return b1 < a1; } };
struct cmpg{ template  bool operator()(const A &a1, const B &b1) { return a1 < b1; } };
#define p_ql(x) priority_queue, cmpl>
#define p_qlp(x, y) priority_queue, vector>, cmpl>
#define p_qg(x) priority_queue, cmpg>
#define p_qgp(x, y) priority_queue, vector>, cmpg>
template bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
#define fo(i,from,to) for(int i=(from),ooo=(from)<(to)?1:-1,oooo=(to)+ooo;i!=oooo;i+=ooo)
#define fol(i,from,to) for(long long i=(from),ooo=(from)<(to)?1:-1,oooo=(to)+ooo;i!=oooo;i+=ooo)
#define foo(i,ooo) for(auto i=ooo.rbegin();i!=ooo.rend();++i)
#define foa(i,from,to) for(int i=(from),ooo=(to);i<=ooo;++i)
#define fos(i,from,to) for(int i=(from),ooo=(to);i>=ooo;--i)

using namespace std;

#ifndef LOCAL
#    define dbg(...) ;
#endif

#define itn int
#define int long long

#ifdef int
#define inf (0x3f3f3f3f3f3f3f3f)
#else
#define inf (0x3f3f3f3f)
#endif

typedef long long ll; typedef unsigned long long ull; typedef pair pii;
const int  dir[8][2]={{0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}}, INF = 0x3f3f3f3f, f_inf = -1044266559, f_INF = -1044266559;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
const int N = 2e6 + 10;
int n, m;

void init()
{
    
}
int dg(int x, int p) 
{
	// dbg(x, p)
	// if(x == 0) return 0;
	// if(x == 2) return 2;
	int num = 0;
	while(x && 2 * p - 1 <= x) {
		x -= 2 * p - 1;
		p--;
		num++;
	}
	// return num + dg(x, p - 1);
	if(x == 0) return num;
	if(x & 1) return num + 1;
	return num + 2;
}
void solve()
{
    int x, y;
	cin >> x >> y;
	int k = sqrt((x + y));
	dbg(k)
	if(k * k != (x + y) || x == 2 || y == 2) {
		cout << "-1\n";
		return;
	}
	int res = dg(x, k);
	cout << res << endl;
}
signed main()
{

#ifdef LOCAL
	freopen("read.in", "r", stdin);
	freopen("write.out", "w", stdout);
#endif

	std::ios::sync_with_stdio(false); std::cin.tie(0); std::cout.tie(0);

	init();

    int t = 0, num2 = 0;

    t = 1;

    if (!t) cin >> t;
    while (t--) {
        //cout<<"Case "<<++num2<<": ";
		solve();
    }
    return 0;

	int num1 = 0;
	//while (scanf("%d", &n) !=-1 && n)
    while (cin >> n && n) {
        //cout<<"Case "<<++num1<<": ";
		solve();
    }
    return 0;

}