Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

5 3
6
4
11

Sample Output 1

3
3
3
2
0

After the first operation, the sequence becomes: 1,2,3,4,5,1.

After the second operation, the sequence becomes: 1,2,3,4.

After the third operation, the sequence becomes: 1,2,3,4,1,2,3,4,1,2,3.

In this sequence, integers 1,2,3,4,5 appear 3,3,3,2,0 times, respectively.

Sample Input 2

10 10
9
13
18
8
10
10
9
19
22
27

Sample Output 2

7
4
4
3
3
2
2
2
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd pair<double,double>
#define X first
#define Y second
#define REP(i,a) for(int i=0;i<a;++i)
#define REPP(i,a,b) for(int i=a;i<b;++i)
#define FILL(a,x) memset(a,x,sizeof(a))
#define	foreach( gg,itit )	for( typeof(gg.begin()) itit=gg.begin();itit!=gg.end();itit++ )
#define	mp make_pair
#define	pb push_back
#define sz(a) int((a).size())
#define all(a)  a.begin(), a.end()
#define	debug(ccc)	cout << #ccc << " = " << ccc << endl;
#define present(c,x) ((c).find(x) != (c).end())
const double eps = 1e-8;
#define EQ(a,b) (fabs((a)-(b))<eps)
inline int max(int a,int b){return a<b?b:a;}
inline int min(int a,int b){return a>b?b:a;}
inline ll max(ll a,ll b){return a<b?b:a;}
inline ll min(ll a,ll b){return a>b?b:a;}
const int mod = 1e9+7;
const int N = 1e6+10;
const ll inf = 1e18;

ll power(ll a,ll n){
	if(n==0){
		return 1;
	}
	ll b = power(a,n/2);
	b = b*b%mod;
	if(n%2) b= b*a%mod;
	return b;
}

int add(int a,int b){ return (a+b)%mod;}
int mul(int a,int b){ return (ll)a*b%mod;}

vector < ll > v;
ll A[N];
ll B[N],ans[N];

int main(){
 // 	freopen("nice.in","r",stdin);
 // freopen("nice.out","w",stdout);
	int n,q;scanf("%d %d",&n,&q);
	v.pb(n);
	REP(i,q){
		ll x;scanf("%lld",&x);
		while(v.size() && v.back()>=x) v.pop_back();
		v.pb(x);
	}
	// REP(i,v.size()) printf("%lld ",v[i]);printf("\n");
	
	A[v.size()-1] = 1;
	for(int i = (int)v.size()-1;i>=1;i--){
		ll cnt = A[i];
		ll p = v[i];
		while(p>v[0]){
			int in = lower_bound(all(v),p) - v.begin();
			in--;
			// printf("%d\n",in);
			A[in] += cnt*(p/v[in]);
			p = p%v[in];
			// printf("%lld\n",p);
		}
		B[p]+=cnt;
	}
	B[v[0]]+=A[0];
	for(int i=v[0];i>0;i--){
		ans[i] = B[i]+ans[i+1];
	}
	REPP(i,1,n+1)	printf("%lld\n",ans[i]);
	return 0;
}