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)

#include <bits/stdc++.h>
#define ll long long
#define fast ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;

ll f( ll a, ll m )
{
    return ( (a % m) * (a % m) ) % m;
}

ll mp[400007];

int main()
{
    ll n,x,m;
    cin>>n>>x>>m;
    vector<ll>v,sum;

    sum.push_back(x);
    v.push_back(x);
    mp[x] = 0;
    ll a= x;
    int l = 0, rep = m;

    for ( int i = 1; i <= m + 2; i++ )
    {
        if ( mp[f(a, m)] or v[0] == f(a, m) )
        {
            rep = i - mp[ f(a, m) ];
            sum.push_back( (sum.back()+f(a, m)) );
            v.push_back( f(a,m) );
            l = mp[f(a, m)];
            break;
        }

        mp[ f(a, m) ] = i;
        sum.push_back( (sum.back()+f(a, m)) );
        v.push_back( f(a,m) );
        a = f(a, m);
    }

    ll xx= 0;

    for( int i=l;i<=l+rep-1;i++ )
        xx+= v[i];

    n--;
    if ( n <= l + rep - 1 )
        cout<< v[n];
    else
    {
        ll blk= (n-l)/rep;
        ll ans= blk*xx;
        cout<<ans+sum[ l + (n - l + rep) % rep ];
    }
}