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 <queue>
typedef long long LL;
typedef std::pair<LL,LL> P;
std::priority_queue<P> Q;
const int N=100005;
LL a[N],cnt[N],ans[N];
void add(int i,LL x,LL y){
	cnt[i]+=x/a[i]*y;
	if (x%a[i]) Q.push(P(x%a[i],y));
}
int main(){
	int n,m,q;
	scanf("%d%d\n",&n,&q);
	a[m=1]=n;
	for (int i=1;i<=q;i++){
		long long x;
		scanf("%lld",&x);
		for (;m && x<=a[m];m--);
		a[++m]=x;
	}
	cnt[m]=1;
	for (int i=m-1;i>=1;i--){
		for (;!Q.empty() && Q.top().first>=a[i];Q.pop()){
			P p=Q.top();
			add(i,p.first,p.second);
		}
		add(i,a[i+1],cnt[i+1]);
	}
	ans[a[1]]+=cnt[1];
	for (;!Q.empty();Q.pop())
		ans[Q.top().first]+=Q.top().second;
	for (int i=a[1];i;i--) ans[i]+=ans[i+1];
	for (int i=1;i<=n;i++) printf("%lld\n",ans[i]);
}