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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define LL long long
#define P pair<int, int>
#define Fengexian cout<<"WZY%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%"<<endl
const LL N = 2e5 + 10;
const LL mod = 998244353;
const LL inf = 0x3f3f3f3f;
const double eps = 1e-9;
using namespace std;
template <typename tp>
inline void read(tp &x)
{
	x = 0; char c = getchar(); bool f = 0;
	for (; c < '0' || c > '9'; f |= (c == '-'), c = getchar()) ;
	for (; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + c - '0', c = getchar()) ;
	if (f) x = -x;
}
LL n,m,q[N],ans[N],times[N];
deque<LL> de;
void solve(LL now)
{
	LL la=q[now];
	while(la>q[1])
	{
		LL pos=lower_bound(q+1,q+m+1,la)-q-1;
		LL flag=la/q[pos];
		la%=q[pos];
		times[pos]+=flag*times[now];
	}
	for(int i=1;i<=la;i++)
		ans[i]+=times[now];
	return;
}
signed main()
{
	read(n),read(m);
	for(int i=1;i<=m;i++) read(q[i]);
	de.push_back(n);
	for(int i=1;i<=m;i++)
	{
		while(!de.empty()&&de.back()>=q[i])
			de.pop_back();
		de.push_back(q[i]);
	}
	m=0;
	while(!de.empty())
		q[++m]=de.front(),de.pop_front();
	times[m]=1;
	for(int i=m;i>=1;i--) solve(i);
	for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}