Given an integer x. Your task is to find out how many positive integers n (1 ≤ n ≤ x) satisfy $$$n ⋅ a^n ≡ b   (mod\;p), where a, b, p$$$ are all known constants.

Input

The only line contains four integers a,b,p,x (2 ≤ p ≤ 10^6+3, 1 ≤ a,b < p, 1 ≤ x ≤ 10^{12}). It is guaranteed that p is a prime.

Output

Print a single integer: the number of possible answers n.

Examples

Input

2 3 5 8


Output

2


Input

4 6 7 13


Output

1


Input

233 233 10007 1


Output

1

Note

In the first sample, we can see that n=2 and n=8 are possible answers.
import bisect
import copy
import decimal
import fractions
import heapq
import itertools
import math
import random
import sys
from collections import Counter,deque,defaultdict
from functools import lru_cache,reduce
from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
def _heappush_max(heap,item):
    heap.append(item)
    heapq._siftdown_max(heap, 0, len(heap)-1)
def _heappushpop_max(heap, item):
    if heap and item < heap[0]:
        item, heap[0] = heap[0], item
        heapq._siftup_max(heap, 0)
    return item
from math import gcd as GCD, modf
read=sys.stdin.read
readline=sys.stdin.readline
readlines=sys.stdin.readlines
 
def LCM(n,m):
    if n or m:
        return abs(n)*abs(m)//math.gcd(n,m)
    return 0

def CRT(lst_r,lst_m):
    r,m=lst_r[0],lst_m[0]
    for r0,m0 in zip(lst_r[1:],lst_m[1:]):
        if (r0,m0)==(-1,0):
            r,m=-1,0
            break
        r0%=m0
        g=math.gcd(m,m0)
        l=LCM(m,m0)
        if r%g!=r0%g:
            r,m=-1,0
            break
        r,m=(r0+m0*(((r-r0)//g)*Extended_Euclid(m0//g,m//g)[0]%(m//g)))%l,l
    return r,m

def Extended_Euclid(n,m):
    stack=[]
    while m:
        stack.append((n,m))
        n,m=m,n%m
    if n>=0:
        x,y=1,0
    else:
        x,y=-1,0
    for i in range(len(stack)-1,-1,-1):
        n,m=stack[i]
        x,y=y,x-(n//m)*y
    return x,y

class MOD:
    def __init__(self,mod):
        self.mod=mod
    
    def Pow(self,a,n):
        a%=self.mod
        if n>=0:
            return pow(a,n,self.mod)
        else:
            assert math.gcd(a,self.mod)==1
            x=Extended_Euclid(a,self.mod)[0]
            return pow(x,-n,self.mod)

    def Build_Fact(self,N):
        assert N>=0
        self.factorial=[1]
        for i in range(1,N+1):
            self.factorial.append((self.factorial[-1]*i)%self.mod)
        self.factorial_inv=[None]*(N+1)
        self.factorial_inv[-1]=self.Pow(self.factorial[-1],-1)
        for i in range(N-1,-1,-1):
            self.factorial_inv[i]=(self.factorial_inv[i+1]*(i+1))%self.mod
        return self.factorial_inv

    def Fact(self,N):
        return self.factorial[N]

    def Fact_Inv(self,N):
        return self.factorial_inv[N]

    def Comb(self,N,K):
        if K<0 or K>N:
            return 0
        s=self.factorial[N]
        s=(s*self.factorial_inv[K])%self.mod
        s=(s*self.factorial_inv[N-K])%self.mod
        return s

a,b,p,x=map(int,readline().split())
if a%p==0 and b%p==0:
    ans=x
elif a%p==0:
    ans=0
elif b%p==0:
    ans=x//p
else:
    ans=0
    MD=MOD(p)
    for i in range(p-1):
        R,M=[i,b*MD.Pow(a,-i)],[p-1,p]
        r,m=CRT(R,M)
        if r==-1:
            continue
        ans+=(x+m-r)//m
print(ans)