The goal is to find $$$\sum^{X-1}_{i=0}A^{i}$$$ $$$ mod $$$ $$$m$$$ , and a simple thought is that we can calculate $$$\frac{A^{X}-1}{A-1}$$$. But when we divide $$$A^{X}-1$$$ by $$$A-1$$$ then $$$mod$$$ $$$m$$$, problem occurs: when we calculate $$$A^{X}-1$$$ , it's already $$$mod$$$ $$$m$$$ ,and there's maybe no reverse to it.

To solve this problem , I use a simple method: store $$$A^i$$$ with two numbers : $$$k$$$ and $$$r$$$ ,satisfying

When we calculate $$$A^{x}$$$,we refresh $$$k$$$ and $$$r$$$ by multiplication ,then refresh $$$k$$$ by modulo $$$m$$$ , and refresh $$$r$$$ when $$$r \ge A-1 $$$ : we do $$$k$$$ $$$+=$$$ $$$r/(A-1)$$$ and $$$r$$$ $$$=$$$ $$$r$$$ % $$$(A-1)$$$ ,like a carry.

We can prove that $$$r=0$$$ when $$$A=2$$$ and $$$r=1$$$ when $$$A=1$$$.if $$$A=1$$$ , we can NOT use $$$k$$$ and $$$r$$$ to store this num for $$$A-1=0$$$ . But this case is easy to solve . The result is obviously $$$X \bmod m$$$ .

Since we get

we can get

So the answer is $$$k_{X}$$$ . Noted that this method is free to change the denominator(in this problem it's $$$A-1$$$).

Code here (GNU C11)

#include<stdio.h>
#define yes printf("Yes\n")
#define no printf("No\n")
//WA but score 500 :)

struct num{
	// store number in form k*deno+r ( 0 <= r < deno)
	// to solve (number^b / deno) mod Mod = ?
	long long k;
	long long r;
};

struct num qtimes(struct num n,long long b,long long deno,long long mod){
	//Warning :deno != 0 
	 if(b>=2){
	 	long long tmpk = (((n.k*n.k%mod)*deno)%mod + (2*n.r*n.k) % mod + n.r*n.r/deno)%mod;
	 	long long tmpr = n.r*n.r % deno;
	 	if (b%2==0) return qtimes({tmpk,tmpr},b/2,deno,mod);
	 	else{
		  	struct num tmpn = qtimes({tmpk,tmpr},b/2,deno,mod);
			long long tk2 = (((n.k*tmpn.k)%mod*deno)%mod + (n.r*tmpn.k+n.k*tmpn.r) % mod + n.r*tmpn.r/deno)%mod;
			long long tr2 = n.r*tmpn.r % deno;
			return {tk2,tr2};  	
		}
	 }
	 else if(b==1) return {(n.k%mod+n.r/deno)%mod,n.r%deno};
	 else if(b==0) return {0,1};
}

int main(){
	long long a,x,m;
	scanf("%lld%lld%lld",&a,&x,&m);
	if (a==1) {
		printf("%lld",x%m); 
		return 0;	
	}
	struct num result = qtimes({1,1},x,a-1,m);
	printf("%lld",result.k+(result.r-1)/(a-1));
	// result = (A^x) ,--> {ans , 1}
}