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Sol

搞个BSGS板子出题人也是很棒棒哦

#include<bits/stdc++.h>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

#define int long long

#define LL long long

#define ull unsigned long long

#define Fin(x) {freopen(#x".in","r",stdin);}

#define Fout(x) {freopen(#x".out","w",stdout);}

using namespace std;

const int MAXN = 1e6 + 10, INF = 1e9 + 10;;

int mod;

const double eps = 1e-9;

template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}

template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}

template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}

template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}

template <typename A> inline void debug(A a){cout << a << '\n';}

template <typename A> inline LL sqr(A x){return 1ll * x * x;}

template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}

template <typename A> A inv(A x) {return fp(x, mod - 2);}

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int G;

map<int, int> mp;

int solve(int x) {//g^ret = x (mod p)

	mp.clear(); int block = ceil(sqrt(mod)), base = fp(G, block);

	for(int i = 0, cur = x; i <= block; i++, mul2(cur, G)) mp[cur] = i;

	for(int i = 1, cur = base; i <= block; i++, mul2(cur, base)) if(mp[cur]) return i * block - mp[cur];

	return 0;

}

/*

int solve(int x) {

	int now = 1;

	for(int i = 0; i<= mod; i++) {

		if(now == x) return i;

		mul2(now, G);

	}

	assert(1 == 2);

}

*/

signed main() {

	//freopen("a.in", "r", stdin);

	G = read(); mod = read();

	int N = read();

	while(N--) {

		int A = read(), B = read();

		cout << fp(G, solve(A) * solve(B)) << '\n';;

	}

	return 0;

}