Given an integer x. Your task is to find out how many positive integers n (1 ≤ n ≤ x) satisfy

where a, b, p are all known constants.

The only line contains four integers a, b, p, x (2 ≤ p ≤ 106 + 3, 1 ≤ a, b < p, 1 ≤ x ≤ 1012). It is guaranteed that p is a prime.

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

In the first sample, we can see that n = 2 and n = 8 are possible answers.

 1 /*
 2 Welcome Hacking
 3 Wish You High Rating
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<ctime>
 9 #include<cstdlib>
10 #include<algorithm>
11 #include<cmath>
12 #include<string>
13 using namespace std;
14 int read(){
15     int xx=0,ff=1;char ch=getchar();
16     while(ch>'9'||ch<'0'){if(ch=='-')ff=-1;ch=getchar();}
17     while(ch>='0'&&ch<='9'){xx=(xx<<3)+(xx<<1)+ch-'0';ch=getchar();}
18     return xx*ff;
19 }
20 long long READ(){
21     long long xx=0,ff=1;char ch=getchar();
22     while(ch>'9'||ch<'0'){if(ch=='-')ff=-1;ch=getchar();}
23     while(ch>='0'&&ch<='9'){xx=(xx<<3)+(xx<<1)+ch-'0';ch=getchar();}
24     return xx*ff;
25 }
26 int mypow(int x,int p,int MOD){
27     int re=1;
28     while(p){
29         if(p&1)
30             re=1LL*re*x%MOD;
31         p>>=1;
32         x=1LL*x*x%MOD;
33     }
34     return re;
35 }
36 int a,b,p;
37 long long x,ans;
38 int main(){
39     //freopen("in","r",stdin);
40     a=read(),b=read(),p=read();
41     x=READ();
42     if(p==2){
43         cout<<x/2+(x%2==1)<<endl;
44         return 0;
45     }
46     long long mul=1LL*p*(p-1);
47     for(int i=0;i<=p-2;i++){
48         int y=1LL*b*mypow(mypow(a,i,p),p-2,p)%p;
49         long long temp=(1LL*i*p*mypow(p,p-3,p-1)%mul+1LL*y*(p-1)*mypow(p-1,p-2,p)%mul)%mul;
50         ans+=x/mul+(x%mul>=temp);
51     }
52     cout<<ans<<endl;
53     return 0;
54 }