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.

Examples

Input

5 3
6
4
11


Output

3
3
3
2
0


Input

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


Output

7
4
4
3
3
2
2
2
0
0
import bisect,sys

input=sys.stdin.readline

N,Q=map(int,input().split())
que=[(N,0)]
for i in range(Q):
    q=int(input())
    que.append((q,i+1))

que.sort(reverse=True)

ext=[]
while que:
    q,id=que.pop()
    if not ext:
        ext.append((q,id))
    else:
        if ext[-1][1]<id:
            if ext[-1][0]==q:
                ext.pop()
            ext.append((q,id))

Q=len(ext)
data=[1]*Q
data[0]=ext[0][0]
edge=[[] for i in range(Q)]
nowext=[ext[0][0]]
for i in range(1,Q):
    q=ext[i][0]
    rest=q
    while True:
        id=bisect.bisect_right(nowext,rest)
        if id==0:
            break
        else:
            edge[id-1].append((i,rest//nowext[id-1]))
            rest%=nowext[id-1]
    nowext.append(ext[i][0])
    data[i]=rest

#print(edge)

dp=[1]*Q
for i in range(Q-2,-1,-1):
    temp=0
    for nv,c in edge[i]:
        temp+=dp[nv]*c
    dp[i]=temp

#print(dp)
#print(data)

minus=[0]*(ext[0][0]+1)
for i in range(Q):
    minus[data[i]]+=dp[i]

base=sum(dp)
for i in range(1,N+1):
    if i-1<=ext[0][0]:
        base-=minus[i-1]
    print(base)
