100 #ABC179E. [ABC179E] Sequence Sum

[ABC179E] Sequence Sum

Score : 500500 points

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

Constraints

  • 1N10101 \leq N \leq 10^{10}
  • 0X<M1050 \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 i=1NAi\displaystyle{\sum_{i=1}^N A_i}.

6 2 1001
1369

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

1000 2 16
6

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

10000000000 10 99959
492443256176507