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}.

Examples

Input

6 2 1001


Output

1369


Input

1000 2 16


Output

6


Input

10000000000 10 99959


Output

492443256176507
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define sz(x) int(x.size())
using namespace std;
typedef long long ll;
const int inf = 1001001001;

int main(){
  ll n;
  ll x,m;
  cin >> n >> x >> m;

  vector<pair<ll,ll>> ord(m);
  ll a = x;
  ll sum = 0;
  for(ll i = 0; i < n; i++){
    if(ord[a].first){
      ll res = n-i-1;
      ll t = i - ord[a].first;
      ll c = res/t;
      ll sumc = (sum - ord[a].second);
      sum += sumc*c;
      i += c*t;
    }else{
          ord[a].first = i;
          ord[a].second = sum;
    }
    sum += a;
    a = (a*a)%m;
  }
  cout << sum << endl;
}
