import sys
#sys.setrecursionlimit(10 ** 6)
INF = float('inf')
#10**20,2**63に変えるのもあり
MOD = 10**9 + 7
MOD2 = 998244353
from collections import defaultdict
def ceil(A,B):
    return -(-A//B)


def egcd(a, b):
    if a == 0:
        return b, 0, 1
    else:
        g, y, x = egcd(b % a, a)
        return g, x - (b // a) * y, y


def crt(b1, m1, b2, m2):
    # 中国剰余定理
    # x ≡ b1 (mod m1) ∧ x ≡ b2 (mod m2) <=> x ≡ r (mod m)
    # となる(r. m)を返す
    # 解無しのとき(0, -1)
    d, p, q = egcd(m1, m2)
    if (b2 - b1) % d != 0:
        return 0, -1
    m = m1 * (m2 // d)  # m = lcm(m1, m2)
    tmp = (b2 - b1) // d * p % (m2 // d)
    r = (b1 + m1 * tmp) % m
    return r, m
def solve():
    def II(): return int(sys.stdin.readline())
    def LI(): return list(map(int, sys.stdin.readline().split()))
    def LC(): return list(input())
    def IC(): return [int(c) for c in input()]
    def MI(): return map(int, sys.stdin.readline().split())
    N,M,P,Q = MI()
    import math
    for q in range(Q):
        X,F = MI()
        X%=P
        G = math.gcd(X,P)
        #G<=P
        #print(G)
        r,m=crt(F,P,0,X)
        if (m == -1):
            print(0)
            continue
        period = P//G
        # G<=P
        #周期
        ans = M//period
        rest_m = M%period
        #print("R:",r,"M:",m)
        
        if(r==0):
            r+=m
        #print(r//X,rest_m)
        if(r//X<=rest_m):
            print(ans+1)
        else:
            print(ans)
    return
solve()