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 <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define per(i,r,l) for(int i=(r);i>=(l);--i)
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
int n,q,tp;ll x,s[maxn],f[maxn],d[maxn];
void add(ll x,ll y){
    int t=lower_bound(s+1,s+tp+1,x)-s;
    if(s[t]>x) --t;
    if(!t) d[1]+=y,d[x+1]-=y;
    else f[t]+=x/s[t]*y,add(x%s[t],y);
}
int main(){
    scanf("%d%d",&n,&q);
    s[++tp]=n;
    rep(i,1,q){
        scanf("%lld",&x);
        while(tp&&s[tp]>x) --tp;
        s[++tp]=x;
    }
    f[tp]=1;
    per(i,tp,2) f[i-1]+=s[i]/s[i-1]*f[i],add(s[i]%s[i-1],f[i]);
    d[1]+=f[1],d[s[1]+1]-=f[1];
    rep(i,1,n) d[i]+=d[i-1],printf("%lld\n",d[i]);
    return 0;
}