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 <iostream>

using namespace std;
typedef long long ll;
bool f[100010];
ll a[100010];
int main(){
    ll i,n,x,m; cin >> n >> x >> m;
    for(i=0;i<m;i++) f[i] = false;
    a[0] = x;
    for(i=1;i<=m;i++) a[i] = a[i - 1]*a[i - 1]%m; 
    int l = -1,r = -1;
    for(i=1;i<=m;i++){
        if(f[a[i]]){
            r = i;
            break;
        }
        f[a[i]]= true;
    }
    for(i=0;i<=m;i++){
        if(a[i]==a[r]){
            l = i; break;
        }
    }
    ll sum = 0,sum1 = 0;
    for(i=0;i<l;i++) sum += a[i];
    for(i=l;i<r;i++) sum1 += a[i];
    n -= l;
    ll ti = n/(r - l),ans = ti*sum1 + sum;
    n -= ti*(r - l);
    for(i=l;i<n + l;i++) ans += a[i];
    cout << ans << endl;
}