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;

long long N,Q,a[100009],p[100009],q[100009];vector<long long>C;

int main(){
	cin>>N>>Q;
	if(Q==0){
		for(int i=1;i<=N;i++)cout<<"1"<<endl;
		return 0;
	}
	for(int i=1;i<=Q;i++) cin>>a[i];
	long long maxn=(1LL<<60);
	for(int i=Q;i>=1;i--){if(maxn>a[i]){maxn=a[i];C.push_back(maxn);}}
	reverse(C.begin(),C.end());
	p[C.size()-1]=1;
	
	for(int i=C.size()-1;i>=0;i--){
		long long cx=C[i];
		while(true){
			long long pos1=upper_bound(C.begin(),C.end(),cx)-C.begin();pos1--;
			if(pos1>=0 && C[pos1]==cx) pos1--;
			if(pos1<0) break;
			//cout<<i<<" "<<p[i]<<" "<<cx<<" "<<pos1<<" "<<C[pos1]<<endl;
			long long I=cx/C[pos1];
			p[pos1]+=I*p[i];
			cx-=I*C[pos1];
		}
		q[N]+=(cx/N)*p[i];
		q[cx%N]+=p[i];
	}
	for(int i=N;i>=1;i--) q[i]+=q[i+1];
	for(int i=1;i<=N;i++) cout<<q[i]<<endl;
	return 0;
}
