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 <bits/stdc++.h>
#define il inline
#define ll long long
const int N=1e5+5;

int n,m=1,q; ll f[N],a[N],ans[N];

il void work(ll c,ll d)
{
    int x=std::upper_bound(a+1,a+m+1,c)-a-1;
    if (!x) ans[1]+=d,ans[c+1]-=d;
    else f[x]+=c/a[x]*d,work(c%a[x],d);
}

int main()
{
    scanf("%d%d",&n,&q),a[m]=n; int i; ll x;
    for ( ; q; q--,a[++m]=x) for (scanf("%lld",&x); x<=a[m]&&m; m--);

    for (f[i=m]=1; i>1; i--) f[i-1]+=a[i]/a[i-1]*f[i],work(a[i]%a[i-1],f[i]);
    for (ans[i=1]+=f[1],ans[a[1]+1]-=f[1]; i<=n; i++) printf("%lld\n",ans[i]+=ans[i-1]);

    return 0;
}