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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(){return +cin[cid++];}
function nextstr(){return cin[cid++];}
function nextbig(){return BigInt(cin[cid++]);}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextssort(n,a){return a?cin.slice(cid,cid+=n).map(a=>+a).sort((a,b)=>b-a):cin.slice(cid,cid+=n).map(a=>+a).sort((a,b)=>a-b);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}
var streams=[];function print(s){streams.push(s);}

var myOut = main();
if(myOut !== undefined)console.log(String(myOut));
if(streams.length)console.log(streams.join("\n"));

function main(){
  var [n,x,m] = nexts(3);
  var set = new Set();
  var ans = 0, xx;
  for(var i = 0; i < n; i++){
    if(set.has(x))break;
    ans += x;
    set.add(x);
    x = x*x%m;
  }
  var sum = 0, xx = x, j = 0;
  for(;i < n;){
    ans += x;
    sum += x;
    x = x*x%m;
    i++,j++;
    if(xx === x)break;
  }
  if(i === n || x === 0)return ans;
  ans += Math.floor((n-i)/j) * sum;
  i += Math.floor((n-i)/j) * j;
  for(;i < n; i++){
    ans += x;
    x = x*x%m;
  }
  return ans;
}