Problem:
Score : 1400 points

Problem Statement

Snuke got an integer sequence from his mother, as a birthday present. The sequence has N elements, and the i-th of them is i.
Snuke performs the following Q operations on this sequence. The i-th operation, described by a parameter q_i, is as follows:

Take the first q_i elements from the sequence obtained by concatenating infinitely many copy of the current sequence, then replace the current sequence with those q_i elements.

After these Q operations, find how many times each of the integers 1 through N appears in the final sequence.

Constraints

1 ≦ N ≦ 10^5

0 ≦ Q ≦ 10^5

1 ≦ q_i ≦ 10^{18}

All input values are integers.

Input

The input is given from Standard Input in the following format:

N Q
q_1
:
q_Q

Output

Print N lines. The i-th line (1 ≦ i ≦ N) should contain the number of times integer i appears in the final sequence after the Q operations.

Sample Input 1

5 3
6
4
11

Sample Output 1

3
3
3
2
0

After the first operation, the sequence becomes: 1,2,3,4,5,1.

After the second operation, the sequence becomes: 1,2,3,4.

After the third operation, the sequence becomes: 1,2,3,4,1,2,3,4,1,2,3.

In this sequence, integers 1,2,3,4,5 appear 3,3,3,2,0 times, respectively.

Sample Input 2

10 10
9
13
18
8
10
10
9
19
22
27

Sample Output 2

7
4
4
3
3
2
2
2
0
0

Solution in Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N, Q = gets.split(" ").map(&:to_i)
q_array = Q.times.map { gets.to_i }

increased = false
prev = 0
started_at = 0
n = (1..N).to_a

q_array.each_cons(3).with_index do |q, i|
  started_at = i if q[1] > prev && !increased
  increased = q[1] > prev
  prev = q[1]
  if q.min == q[1]
    n = n.cycle(q[1] / n.size + 1).to_a[0...q[1]]
  end
end

q_array[started_at..-1].each do |q|
  n = n.cycle(q / n.size + 1).to_a[0...q]
end

counts = n.group_by { |i| i }

1.upto(N) do |i|
  puts counts.key?(i) ? counts[i].size : 0
end