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;
const int N=1e5+10;
#define ll long long
ll stk[N],f[N],ans[N];
int n,top,q;
void solve(ll x,ll v){
    int k=upper_bound(stk+1,stk+top+1,x)-stk-1;
    if(!k){
        ans[1]+=v;ans[x+1]-=v;
        return ;
    }
    f[k]+=v*(x/stk[k]);solve(x%stk[k],v);
}
int main(){
    scanf("%d%d",&n,&q);stk[++top]=n;
    for(int i=1;i<=q;i++){
        ll x;scanf("%lld",&x);
        while(x<=stk[top]&&top)--top;
        stk[++top]=x;
    }
    f[top]=1;
    for(int i=top-1;i;i--)
        f[i]+=f[i+1]*(stk[i+1]/stk[i]),solve(stk[i+1]%stk[i],f[i+1]);
    ans[1]+=f[1];ans[stk[1]+1]-=f[1];
    for(int i=1;i<=n;i++)printf("%lld\n",ans[i]+=ans[i-1]);
}
