#abc179e. [abc179_e]Sequence Sum

[abc179_e]Sequence Sum

Problem Statement

Let us denote by f(x,m)f(x, m) the remainder of the Euclidean division of xx by mm.

Let AA be the sequence that is defined by the initial value A1=XA_1=X and the recurrence relation An+1=f(An2,M)A_{n+1} = f(A_n^2, M). Find displaystylesumi=1NAi\\displaystyle{\\sum_{i=1}^N A_i}.

Constraints

  • 1leqNleq10101 \\leq N \\leq 10^{10}
  • 0leqX<Mleq1050 \\leq X < M \\leq 10^5
  • All values in input are integers.

Input

Input is given from Standard Input in the following format:

NN XX MM

Output

Print displaystylesumi=1NAi\\displaystyle{\\sum_{i=1}^N A_i}.


Sample Input 1

6 2 1001

Sample Output 1

1369

The sequence AA begins 2,4,16,256,471,620,ldots2,4,16,256,471,620,\\ldots Therefore, the answer is 2+4+16+256+471+620=13692+4+16+256+471+620=1369.


Sample Input 2

1000 2 16

Sample Output 2

6

The sequence AA begins 2,4,0,0,ldots2,4,0,0,\\ldots Therefore, the answer is 66.


Sample Input 3

10000000000 10 99959

Sample Output 3

492443256176507