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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=100010;
ll v[N],f[N],ans[N];
inline ll gi() {
    ll x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
int main() {
    int n,Q,tp=1;
    cin>>n>>Q,v[1]=n;
    for(int i=1;i<=Q;i++) {
	ll x=gi();
	while(x<=v[tp]) --tp;
	v[++tp]=x;
    }
    f[tp]=1;
    for(int i=tp;i;i--) {
	ll p=v[i];
	while(p>v[1]) {
	    int t=upper_bound(v+1,v+i,p)-v-1;
	    f[t]+=p/v[t]*f[i],p%=v[t];
	}
	ans[p]+=f[i];
    }
    for(int i=n;i;i--) ans[i]+=ans[i+1];
    for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
    return 0;
}