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;

using ll=long long int;
ll n,x,m;

int main()
{
    cin>>n>>x>>m;
    vector<ll>cnt(m,0);
    queue<ll>rem;
    ll idx=1,sum=x,ans=0,next=(x*x)%m,loop=0;
    cnt[x]=idx;
    rem.push(x);
    while(1)
    {
        if (cnt[next] != 0) 
            break;
        cnt[next] = ++idx;
        rem.push(next);
        sum += next, next = (next * next) % m;
    }
    for (auto i = 1; i < cnt[next]; i++) 
    {
        ans += rem.front();
        rem.pop();
    }

    loop = rem.size(), n -= (cnt[next] - 1),
    ans += (n / loop) * (sum - ans);
    for (auto i = 0; i < n % loop; i++) 
    {
        ans += rem.front();
        rem.pop();
    }
    cout << ans << endl;
}

