Problem:
Score : 500 points

Problem Statement

Let us denote by f(x, m) the remainder of the Euclidean division of x by m.

Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M).
Find \displaystyle{\sum_{i=1}^N A_i}.

Constraints

1 \leq N \leq 10^{10}

0 \leq X < M \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X M

Output

Print \displaystyle{\sum_{i=1}^N A_i}.

Sample Input 1

6 2 1001

Sample Output 1

1369

The sequence A begins 2,4,16,256,471,620,\ldots Therefore, the answer is 2+4+16+256+471+620=1369.

Sample Input 2

1000 2 16

Sample Output 2

6

The sequence A begins 2,4,0,0,\ldots Therefore, the answer is 6.

Sample Input 3

10000000000 10 99959

Sample Output 3

492443256176507

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

 //----AUTHOR:kkdrummer----/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
//#include <boost/multiprecision/cpp_int.hpp> 
//using namespace boost::multiprecision; 
using namespace __gnu_pbds;
using namespace std;
typedef    long long                	 ll;
typedef    long double              	 ld;
typedef    unordered_set<ll>         	 usll;
typedef    unordered_multiset<ll>    	 umsll;
typedef    multiset<ll>              	 msll;
typedef    set<ll>                  	 sll;
typedef    vector<ll>               	 vll;
typedef    pair<ll,ll>              	 pll;
typedef    vector<pll>              	 vpll;
typedef    priority_queue<ll>       	 pqll;
typedef    vector<int>              	 vi;
typedef    set<int>                 	 si;
typedef    multiset<int>            	 msi;
typedef    unordered_multiset<int>  	 umsi;
typedef    unordered_set<int>       	 usi;
typedef    pair<int,int>               	 pi;
typedef    vector<pi>                	 vpi;
typedef    priority_queue<int>       	 pqi;
typedef    tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>   ind_si;
typedef    tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>   ind_sll;
typedef    tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>   ind_msi;
typedef    tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update>   ind_msll;
#define    in          		insert
#define    fi          		first
#define    se          		second
#define    pb   	    	push_back
#define    mp   	   		make_pair
#define    be   	   		begin
#define    en   	   		end
#define    itr  	   		iterator
#define    io  	       		ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define    mo   	   		998244353
#define    inf  	   		8222372026854775807
#define    ninf   	   		-inf
#define    ima   	   		1147483647
#define    imi   	   		-ima
#define    oncnt       		__builtin_popcount
#define    zerobegin   		__builtin_clz
#define    zeroend    		__builtin_ctz
#define    parity      		__builtin_parity
#define    all(x)      		x.be(),x.en()
#define    eps         		1e-9
#define    coutd       		cout<<setprecision(10)<<fixed
#define    mems(dp,x)  		memset(dp,x,sizeof(dp))
#define    fbo         		find_by_order
#define    ook    	   		order_of_key
#define    upb    	   		upper_bound
#define    lowb  	   		lower_bound 
#define    lte(v,x)    		(upb(all(v),x)-v.be())
#define    gte(v,x)    		(v.end()-lowb(all(v),x))
#define    gt(v,x)     		(v.en()-upb(all(v),x))
#define    lt(v,x)     		(lowb(all(v),x)-v.be())
const      ld PI= 			3.1415926535897932384626433832792884197169399375105820974944;
inline ll modpow(ll x,ll n){if(n==0)return 1;if(x==0)return 0;if(n==1)return(x%mo);ll u=(modpow(x,n/2));u=(u*u)%mo;if(n%2!=0)u=(u*x%mo)%mo;return u;}
inline ll modinv(ll x){return modpow(x,mo-2);}
inline ll mmul(ll a,ll b){if(a>=mo)a=a%mo;if(b>=mo)b=b%mo;if(a*b>=mo)return(a*b)%mo;return(a*b);}
inline ll madd(ll a, ll b){if(a>=mo)a=a%mo;if(b>=mo)b=b%mo;if(a+b>=mo)return(a+b)%mo;return(a+b);}
inline ll msub(ll a, ll b){if(a>=mo)a=a%mo;if(b>=mo)b=b%mo;return(((a-b)%mo+mo)%mo);}
inline ll mdiv(ll a,ll bb){if(a>=mo)a=a%mo;ll b=modinv(bb);if(b>=mo)b=b%mo;if(a*b>=mo)return(a*b)%mo;return(a*b);}
inline ll gcd(ll a,ll b){return __gcd(a,b);}
inline ll lcm(ll a,ll b){return (a*b)/gcd(a,b);}
 

int a[1000001]={0};


int main()
{               io 
int testcases=1;             //          cin>>testcases;
while(testcases--)
{

ll n,x,m;
cin>>n>>x>>m;

if(n==1)
{
	cout<<x;
	return 0;
}

if(x==0)
{
	cout<<0;
	return 0;
}

vll v;
map<ll,int> s;
s[x]=0;
v.pb(x);
int i=1;
ll pr=x;
ll size;
while(true)
{
	ll c=(pr*pr)%m;
	pr=c;
	if(s.find(c)!=s.en())
	{
		size=i-s[c];
		break;
	}
	v.pb(c);
	s[c]=i;
	i++;
}

vi ne;
for(int i=0;i<size;i++)
{
	ne.pb(v.back());
	v.pop_back();
}

int bs=v.size();
reverse(all(ne));



for(int i=1;i<v.size();i++)
v[i]+=v[i-1];
for(int i=1;i<ne.size();i++)
ne[i]+=ne[i-1];

if(n<=bs)
{
	cout<<v[n-1];
	return 0;
}



ll c=v[bs-1];
n-=bs;
if(size>0)
c+=(n/size)*ne[size-1];
if(n%size!=0&&size>0)
c+=ne[(n-1)%size];
cout<<c;









































}return 0;}