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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program sequential_operations_on_sequence
  implicit none
  integer :: n, q, i, j, x, u = 1, l, r, m
  integer(8) :: p(100000) = 0_8, k, t(100000) = 0_8, ans(100001) = 0_8
  read(*,*) n, q
  p(1) = int(n,8)
  do i = 1, q
    u = u+1
    read(*,*) p(u)
    do while (u > 1 .and. p(u-1) >= p(u))
      p(u-1) = p(u)
      u = u-1
    end do
  end do
  t(u) = 1_8
  do i = u, 1, -1
    j = i
    k = p(i)
    do
      r = j
      l = 0
      do while (r-l > 1)
        m = (l+r)/2
        if (p(m) > k) then
          r = m
        else
          l = m
        end if
      end do
      if (l == 0) then
        x = int(k,4)
        ans(1) = ans(1)+t(i)
        ans(x+1) = ans(x+1)-t(i)
        exit
      else
        j = l
        t(j) = t(j)+t(i)*(k/p(j))
        k = mod(k,p(j))
      end if
    end do
  end do
  write(*,'(i0)') ans(1)
  do i = 2, n
    ans(i) = ans(i)+ans(i-1)
    write(*,'(i0)') ans(i)
  end do
end program sequential_operations_on_sequence