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 F first
#define S second
#define X real()
#define Y imag()
using namespace std;
typedef long long ll;
typedef long double ld;

ll v[101010];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	ll n,q;
	cin>>n>>q;
	vector<ll> qs;
	qs.push_back(n);
	for (int i=0;i<q;i++){
		ll qq;
		cin>>qq;
		while (qs.size()>0&&qs.back()>=qq) qs.pop_back();
		qs.push_back(qq);
	}
	map<ll, ll> qq;
	qq[qs.back()]=1;
	for (int i=(int)qs.size()-2;i>=0;i--){
		ll t=qs[i];
		//cout<<t<<endl;
		while ((*qq.rbegin()).F>t){
			ll k=(*qq.rbegin()).F;
			ll a=(*qq.rbegin()).S;
			//cout<<k<<" "<<a<<endl;
			qq.erase(k);
			ll wt=k/t;
			qq[t]+=a*wt;
			if (k%t>0){
				qq[k%t]+=a;
			}
		}
	}
	for (auto a:qq){
		v[a.F]+=a.S;
	}
	for (int i=n;i>=0;i--){
		v[i]+=v[i+1];
	}
	for (int i=1;i<=n;i++){
		cout<<v[i]<<"\n";
	}
}