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
/*
17/10/2017
*/
#include <bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;
int main()
{
ll n,x,m;
cin>>n>>x>>m;
ll ans=0;
ll k=x%m;
ans+=k;
n--;
vector<ll> vc;
vc.push_back(ans);
map<ll,ll> mp;
mp[k]=1;
int cnt=0;
while(n>0)
{
k=(k*k)%m;
//cnt++;
//if(k==10000)
//{
//cout<<cnt<<endl;
//}
if(mp.count(k))
{
ll l=n/(vc.size()-mp[k]+1);
if(mp[k]==1)
{
ans+=(vc[vc.size()-1])*l;
}
else
{
ans+=(vc[vc.size()-1]-vc[mp[k]-2])*l;
}
n%=(vc.size()-mp[k]+1);
//cout<<ans<<endl;
break;
}
else
{
n--;
}
ans+=k;
vc.push_back(ans);
mp[k]=vc.size();
}
k=vc[vc.size()-1]-vc[vc.size()-2];
while(n>0)
{
k=(k*k)%m;
ans+=k;
n--;
}
cout<<ans;
}