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
/*
 * @Author: BilyHurington
 * @Date: 2020-09-15 13:22:15
 * @LastEditors: BilyHurington
 * @LastEditTime: 2020-09-15 13:45:06
 */
#include<bits/stdc++.h>
using namespace std;
int n,q,m;
long long st[100010],num[100010],cnt[100010];
void add_num(int l,int r,long long t){num[l]+=t;num[r+1]-=t;}
void calc(int nxt,long long now,long long ti){
    cnt[nxt]+=ti*(now/st[nxt]);now%=st[nxt];
    int x=upper_bound(st+1,st+m+1,now)-st-1;
    if(!x) add_num(1,now,ti);
    else calc(x,now,ti);
}
int main(){
    scanf("%d %d",&n,&q);
    st[m=1]=n;long long x;
    for(int i=1;i<=q;i++){
        scanf("%lld",&x);
        while(m&&x<=st[m]) m--;
        st[++m]=x;
    }
    cnt[m]=1;
    for(int i=m;i>1;i--) calc(i-1,st[i],cnt[i]);
    for(int i=1;i<=n;i++) num[i]+=num[i-1];
    for(int i=1;i<=st[1];i++) num[i]+=cnt[1];
    for(int i=1;i<=n;i++) printf("%lld\n",num[i]);
    return 0;
}