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 int maxn = 3e5 + 5;

ll n;
int x, p;
int a[maxn], vis[maxn];
ll b[maxn];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	scanf("%lld%d%d", &n, &x, &p);
	a[1] = x, vis[x] = 1;
	int l, r, cnt = 1;
	while(1){
		a[cnt+1] = 1ll * a[cnt] * a[cnt] % p;
		++cnt;
		if(vis[a[cnt]]){
			l = vis[a[cnt]], r = cnt - vis[a[cnt]];
			break;
		}
		vis[a[cnt]] = cnt;
	}
	for(int i = 1; i <= cnt; ++i) b[i] = b[i-1] + 1ll * a[i];
	
	cout << (b[cnt-1] - b[l-1]) * ((n - l + 1) / r) + b[(n-l+1)%r+l-1];
}