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 (ll i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int main() {
  ll n, x, m, logn, ans = 0;
  cin >> n >> x >> m;
  logn = ceil(log2(n)) + 1;
  vector<vector<ll>> nex(logn, vector<ll>(m)), sum(logn, vector<ll>(m));

  rep(i, m) {
    nex[0][i] = i * i % m;
    sum[0][i] = i;
  }

  rep(i, logn - 1) rep(j, m) {
    nex[i + 1][j] = nex[i][nex[i][j]];
    sum[i + 1][j] = sum[i][nex[i][j]] + sum[i][j];
  }

  rep(i, logn) if (n & (1ll << i)) ans += sum[i][x], x = nex[i][x];
  cout << ans << endl;
  return 0;
}