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: Runtime Error)

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
 
using namespace std;

typedef long long ll;

inline ll read()
{
	ll x=0;char ch=getchar();
	while(ch<'0' || '9'<ch)ch=getchar();
	while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
	return x;
}
 
const int N=1e5+9;

ll n,q;
ll a[N],top=0;
ll cnt[N];
ll ans[N];

int main()
{
	n=read();
	q=read();

	a[0]=n;
	for(int i=1,x;i<=q;i++)
	{
		x=read();
		while(top>=0 && a[top]>=x)top--;
		a[++top]=x;
	}

	cnt[top]=1;
	for(int i=top;i;i--)
	{
		ll x=a[i];
		while(x>=a[0])
		{
			int pos=upper_bound(a,a+i,x)-a-1;
			cnt[pos]+=x/a[pos]*cnt[i];
			x%=a[pos];
		}
		ans[x]+=cnt[i];
	}

	ans[a[0]]+=cnt[0];
	for(int i=n;i;--i)
		ans[i-1]+=ans[i];
	for(int i=1;i<=n;++i)
		printf("%lld\n",ans[i]);
	return 0;
}