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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;
int n,m,top,nn;
ll a[100005],s[100005],ans[100005];
map<ll,ll> mp;
void dfs(ll x){
	if(x<=n) return;
	ll w=mp[x];int l=1,r=top,mid,pos;
	while(l<=r){
		mid=l+r>>1;
		if(s[mid]<x) pos=mid,l=mid+1;
		else r=mid-1;
	}
	if(mp.find(s[pos])==mp.end()) mp[s[pos]]=0;
	ll rst=x%s[pos];
	if(mp.find(rst)==mp.end()) mp[rst]=0;
	mp[s[pos]]+=x/s[pos]*w;
	mp[rst]+=w;
	dfs(s[pos]);
	dfs(rst);
}
int main(){
	scanf("%d%d",&n,&m);
	a[1]=n;
	for(int i=1;i<=m;i++) scanf("%lld",&a[i+1]);
	m++;
	for(int i=1;i<=m;i++){
		while(top&&a[i]<=s[top]) top--;
		s[++top]=a[i];
	}
	if(s[1]<n) nn=n,n=s[1];
	mp[s[top]]=1;
	dfs(s[top]);
	ll sum=0;
	for(int i=n;i;i--){
		if(mp.find(i)!=mp.end()) sum+=mp[i];
		ans[i]=sum;
	}
	for(int i=1;i<=nn;i++) printf("%lld\n",ans[i]);
	return 0;
}