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 <iostream>
#include <algorithm>
#define N 100010
#define ll long long
using namespace std;
ll n,m,cnt,a[N],f[N],Q[N],b[N];
void work(ll x,ll y)
{
	ll p=upper_bound(Q,Q+cnt+1,x)-Q-1;
	if (!p) b[1]+=y,b[x+1]-=y; else f[p]+=y*(x/Q[p]),work(x%Q[p],y);
}
int main()
{
	scanf("%lld%lld",&n,&m);
	for (ll i=1;i<=m;i++) scanf("%lld",&a[i]);
	if (!m){ for (ll i=1;i<=n;i++) printf("1\n"); return 0; }
	Q[++cnt]=a[m]; for (ll i=m-1;i;i--) if (a[i]<Q[cnt]) Q[++cnt]=a[i];
	if (n<Q[cnt]) Q[++cnt]=n; 
	reverse(Q+1,Q+cnt+1),f[cnt]=1;
	for (ll i=cnt;i>1;i--) f[i-1]+=f[i]*(Q[i]/Q[i-1]),work(Q[i]%Q[i-1],f[i]);
	for (ll i=1;i<=n;i++) b[i]+=b[i-1];
	for (ll i=1;i<=Q[1];i++) b[i]+=f[1];
	for (ll i=1;i<=n;i++) printf("%lld\n",b[i]);
}