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<cstdio>
#include<algorithm>
#include<map>
using namespace std;

typedef long long ll;
const int MAXN=1e5+5;

int n,m;
ll ans[MAXN],q[MAXN];
map<ll,ll> a;
map<ll,ll>::iterator it;

int main(){

    scanf("%d%d",&n,&m);
    if(m==0){
        for(int i=1;i<=n;i++) puts("1");
        return 0;
    }
    for(int i=1;i<=m;i++) scanf("%lld",&q[i]);
    a[q[m]]++; q[0]=n;
    for(int i=m-1;i>=0;i--)
        while(a.rbegin()->first>q[i]){
            it=a.end(); it--;
            a[it->first%q[i]]+=it->second;
            a[q[i]]+=it->second*(it->first/q[i]);
            a.erase(it);
        }
    for(it=a.begin();it!=a.end();it++) ans[it->first]+=it->second;
    for(int i=n;i>=1;i--) ans[i]+=ans[i+1];
    for(int i=1;i<=n;i++)
        printf("%lld\n",ans[i]);

    fclose(stdin);
    fclose(stdout);
}
