Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
const int N=3e5+5;
using namespace std;
typedef long long ll;
typedef map<ll,ll>::iterator mit;
int n,q;
ll p[N];
map<ll,ll> mp;
ll cnt[N];
int main()
{
    scanf("%d%d",&n,&q);
    if(q==0){
        for(int i=1;i<=n;i++)
            puts("1");
        return 0;
    }
    for(int i=1;i<=q;i++)
        scanf("%lld",&p[i]);
    mp[p[q]]++;
    p[0]=n;
    for(int i=q-1;i>=0;i--){
        while(mp.rbegin()->first>p[i]){
            mit it=mp.end();
            it--;
            mp[it->first%p[i]]+=it->second;
            mp[p[i]]+=it->second*(it->first/p[i]);
            mp.erase(it);
        }
    }
    for(mit it=mp.begin();it!=mp.end();it++)
        cnt[it->first]+=it->second;
    for(int i=n;i>=1;i--)
        cnt[i]=cnt[i]+cnt[i+1];
    for(int i=1;i<=n;i++)
        printf("%lld\n",cnt[i]);
}