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
#include<iostream>
#include<cstdio>
#include<deque>
#include<algorithm>
#define LL long long
#define reint register int
using namespace std;
const int N=100005;
deque<LL>que;
LL b,s[N],cot[N],ans[N];int n,m,cnt;
void push(LL x){
    while(!que.empty()&&que.back()>=x)que.pop_back();
    que.push_back(x);return;
}
int main(){
    scanf("%d%d",&n,&m);push(n);
    for(reint i=1;i<=m;i++)
    scanf("%lld",&b),push(b);
    while(!que.empty())
        s[++cnt]=que.front(),que.pop_front();
    cot[cnt]=1;
    for(reint i=cnt-1;~i;i--){
        LL len=s[i+1];
        while(len>s[1]){
            reint pos=upper_bound(s+1,s+i+1,len)-s-1;
            cot[pos]+=len/s[pos]*cot[i+1];
            len%=s[pos];
        }ans[len]+=cot[i+1];
    }
    for(reint i=s[1];i;i--)ans[i]+=ans[i+1];
    for(reint i=1;i<=n;i++)printf("%lld\n",ans[i]);
    return 0;
}