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>
using namespace std;
typedef long long ll;
const ll MAXN=1e5+10;
ll N,X,M;
set<ll> s;
map<ll,ll> m;
ll Add[MAXN];
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin>>N>>X>>M;
    ll A,i;
    for(A=X,i=1;s.find(A)==s.end();A=A*A%M,i++)
        Add[i]=Add[i-1]+A,s.insert(A),m[A]=i;
    ll Pos=m[A];
    ll T=i-Pos,Sum=Add[i-1]-Add[Pos-1];
    ll Ans=(N-Pos+1)/T*Sum+Add[(N-Pos+1)%T+Pos-1];
    cout<<Ans<<endl;
    return 0;
}