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 <algorithm>
using namespace std;
typedef long long ll;
const int N = 100100;
ll q[N],f[N],tag[N],x;
int n,Q,tail;
void solve(ll d,ll v) {
    if(d==0) return ;
    int j=upper_bound(q+1,q+tail+1,d)-q-1;
    if(j==0) tag[1]+=v,tag[d+1ll]-=v;
    else f[j]+=d/q[j]*v,solve(d%q[j],v);
}
int main() {
    scanf("%d%d",&n,&Q);
    q[++tail]=(ll)n;
    for(int i=1;i<=Q;i++) {
        scanf("%lld",&x);
        while(tail&&q[tail]>=x) tail--;
        q[++tail]=x;
    }
    f[tail]=1;
    for(int i=tail;i>1;i--) {
        f[i-1]+=q[i]/q[i-1]*f[i];
        solve(q[i]%q[i-1],f[i]);
    }
    tag[1]+=f[1],tag[q[1]+1ll]-=f[1];
    for(int i=1;i<=n;i++) tag[i]+=tag[i-1],printf("%lld\n",tag[i]);
}