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<bits/stdc++.h>
#define int long long
#define N 100005
using namespace std;
int t[N],n,m;
int a[N],top,b[N];
int bi(int x)
{
	int l=1,r=top;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(b[mid]>x)r=mid-1;
		else l=mid+1;
	}
	return r;
}
signed main()
{
	scanf("%lld%lld",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int tmp;scanf("%lld",&tmp);
		while(top&&tmp<=b[top])top--;
		b[++top]=tmp;
	}
	t[top]=1;
	for(int i=top;i>=1;i--)
	{
		int k=b[i];
		if(i!=1)
		{
			t[i-1]+=b[i]/b[i-1]*t[i];
			k=b[i]-b[i]/b[i-1]*b[i-1];
		}
		while(k>b[1])
		{
			int p=bi(k);
			t[p]+=k/b[p]*t[i];
			k%=b[p];
		}
		a[1]+=t[i];a[k+1]-=t[i];
	}
	int now=0;
	for(int i=1;i<=n;i++)
	{
		now+=a[i];
		printf("%lld\n",now);
	}
	return 0;
}