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>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;

ll n, x, p;
ll a[maxn], sum[maxn], vis[maxn];
ll ans;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cin >> n >> x >> p;
	a[1] = sum[1] = x, vis[x] = 1;
	for(int i = 2; i <= n; ++i){
		a[i] = ((a[i-1] % p) * (a[i-1] % p)) % p;
		if(!a[i]) break;
		if(vis[a[i]]){
			ll seg = i - vis[a[i]];
			ans += (n - i + 1) / seg * (sum[i-1] - sum[vis[a[i]]-1]) + sum[vis[a[i]]+(n-i+1)%seg-1] - sum[vis[a[i]]-1];
			break;
		}
		ans = sum[i] = sum[i-1] + a[i]; 
		vis[a[i]] = i;
	}
	cout << ans;
}