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<bits/stdc++.h>

using namespace std;
const int m=1e5+60;
long long pom[m];
long long odp[m];

int main(){
  ios_base::sync_with_stdio(0);
  int n,q;
  cin>>n>>q;
  vector<long long> v;
  v.push_back(n);
  for(int i=0;i<q;i++) {
    long long in;
    cin>>in;
    while(v.back()>=in && !v.empty())v.pop_back();
    v.push_back(in);
  }
 // cout<<"co jest" << v.size()-1<<endl;
  int siz=v.size()-1;
  pom[siz]=1;
  int p;
  for(int i=siz;i>0;i--) {
    long long k=v[i];
    while(k>v[0]){
      p=lower_bound(v.begin(),v.begin()+i,k)-v.begin()-1;
      pom[p]+=pom[i]*(k/v[p]);
      k=k%v[p];
    }
    odp[k-1]+=pom[i];
  }
  odp[v[0]-1]+=pom[0];
  for(int i=n-1;i>=0;i--) {
    odp[i]+=odp[i+1];
  }
  for(int i=0;i<n;i++) {
    cout<<odp[i]<<endl;
  }
}
