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>
using namespace std;

using ll=long long;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	for(int n,q;cin>>n>>q && n|q;){
		vector<ll> qs(1,n);
		for(int i=0;i<q;i++){
			ll x; cin>>x;
			while(qs.size() && qs.back()>=x)
				qs.pop_back();
			qs.push_back(x);
		}
		vector<ll> cs(qs.size());
		cs.back()=1;
		vector<ll> res(n+1);
		for(int i=qs.size();i--;){
			for(;;){
				int j=upper_bound(begin(qs),begin(qs)+i,qs[i])-begin(qs)-1;
				if(j<0) break;
				cs[j]+=qs[i]/qs[j]*cs[i];
				qs[i]%=qs[j];
			}
			res[0]+=cs[i];
			res[qs[i]]-=cs[i];
		}
		for(int i=0;i<n;i++){
			cout<<res[i]<<endl;
			res[i+1]+=res[i];
		}
	}
}
