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
// E - Sequence Sum
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
pair<ll,ll> divmod(ll p,ll m){return {p/m, p%m};}

int main(){
	ll n; int x, m; cin>>n>>x>>m;
	vector<int> X(m, -1);
	vector<int> P;
	ll sum_p = 0;
	while(X[x] < 0){ // preset
		X[x] = P.size(); // pre length
		P.push_back(sum_p); // pre sum_p
		sum_p += x; // now sum_p
		x = (ll)x*x % m;
	}
	ll cyc_times, nxt_len;
	tie(cyc_times, nxt_len) = divmod(n - X[x], P.size() - X[x]);
	ll cyc = (sum_p - P[X[x]]) * cyc_times;
	ll remain = P[X[x] + nxt_len];
	cout<< cyc + remain <<endl;
}
