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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  ll n, x, m;
  cin >> n >> x >> m;
  vector<int> id(m,-1);
  vector<int> a;
  int len = 0;
  ll tot = 0;
  while (id[x] == -1) {
    a.push_back(x);
    id[x] = len;
    len++;
    tot += x;
    x = (x*x)%m;
  }

  int c = len-id[x];
  ll s = 0;
  for (int i = id[x]; i < len; ++i) s += a[i];

  ll ans = 0;
  if (n <= len) {
    rep(i,n) ans += a[i];
  } else {
    ans += tot;
    n -= len;
    ans += s*(n/c);
    n %= c;
    rep(i,n) ans += a[id[x]+i];
  }
  cout << ans << endl;
  return 0;
}
