#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
using namespace std;
// using namespace __gnu_pbds;
// template<class T>
// using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T, class U>
void ckmin(T &a, U b)
{
if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
if (a < b) a = b;
}
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
const int MAXN = 100013;
int T, N, Q;
ll C;
pll coor[MAXN];
int ccw(pll a, pll b, pll c)
{
a.fi -= b.fi; a.se -= b.se;
c.fi -= b.fi; c.se -= b.se;
__int128_t cp = (__int128_t) a.fi * c.se - a.se * c.fi;
if (cp > 0) return 1;
if (cp < 0) return -1;
return 0;
//idk if this casts properly check it out
}
pll p;
bool ask(int i)
{
return ccw(p, coor[i], coor[i + 1]) <= 0;
}
int32_t main()
{
cout << fixed << setprecision(12);
cerr << fixed << setprecision(4);
ios_base::sync_with_stdio(false); cin.tie(0);
cin >> T >> N;
FOR(i, 0, N)
{
cin >> coor[i].fi >> coor[i].se;
}
coor[N] = coor[0];
cin >> Q;
while(Q--)
{
cin >> p.fi >> p.se;
if (T)
{
p.fi ^= (C * C * C);
p.se ^= (C * C * C);
}
// FOR(i, 0, N)
// {
// cerr << ask(i) << ' ';
// }
// cerr << endl;
//ask (0, n/2). if you get two 1's, you win. if you get two 0's you lose.
//otherwise you get 01. then you binary search.
bool lt = ask(0), rt = ask(N/2), ans;
if (lt == rt)
{
ans = lt;
}
else
{
int lpos, rpos;
int lo = 0, hi = N/2 - 1;
while(hi > lo)
{
int mid = (hi + lo + 1) >> 1;
if (ask(mid) == lt) lo = mid;
else hi = mid - 1;
}
lpos = lo; //hi is the last one that's = lo.
lo = N/2; hi = N - 1;
while(hi > lo)
{
int mid = (hi + lo + 1) >> 1;
if (ask(mid) == rt) lo = mid;
else hi = mid - 1;
}
rpos = lo;
//so lpos+1....rpos are rt.
int c = rpos - lpos;
ans = (c < N / 2 && !rt) || (c > N / 2 && rt);
}
cout << (ans ? "DA" : "NE") << '\n';
if (ans) C++;
}
}
