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<cstring>
#include<algorithm>
#include<map>
#define SF scanf
#define PF printf
#define MAXN 300010
#define INF 0x3FFFFFFF
#define MOD 998244353
using namespace std;
typedef long long ll;
typedef map<ll,ll>::iterator mit;
int n,q;
ll qx[MAXN];
map<ll,ll> mp;
ll used[MAXN];
int main(){
	SF("%d%d",&n,&q);
	if(q==0){
		for(int i=1;i<=n;i++)
			PF("1\n");
		return 0;	
	}
	for(int i=1;i<=q;i++)
		SF("%lld",&qx[i]);
	mp[qx[q]]++;
	qx[0]=n;
	for(int i=q-1;i>=0;i--){
		while(mp.rbegin()->first>qx[i]){
			mit it=mp.end();
			it--;	
			mp[it->first%qx[i]]+=it->second;
			mp[qx[i]]+=it->second*(it->first/qx[i]);
			mp.erase(it);
		}
	}
	mit it=mp.begin();
	for(;it!=mp.end();it++)
		used[it->first]+=it->second;
	for(int i=n;i>=1;i--)
		used[i]=used[i]+used[i+1];
	for(int i=1;i<=n;i++)
		PF("%lld\n",used[i]);
}