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.

Examples

Input

5 3
6
4
11


Output

3
3
3
2
0


Input

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


Output

7
4
4
3
3
2
2
2
0
0
#include<bits/stdc++.h>
using namespace std;
 
#define ll long long
#define pb push_back
 
const int maxn = 2e5 + 20;

ll a[maxn] , t[maxn];

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	int n , m;
	cin >> n >> m;

	a[0] = n;
	for(int i = 1; i <= m; i++)
	{
		cin >> a[i];
		int j = i - 1;
		while(j >= 0 && a[i] <= a[j])
			j--;

		a[j + 1] = a[i];
		m -= (i - j - 1);
		i = j + 1;
	}

	map<ll , ll> mp;
	mp[a[m]]++;

	for(int i = m; i > 0; i--)
		while(1)
		{
			ll x = mp.rbegin() -> first , y = mp.rbegin() -> second;
			if(x > a[i - 1])
			{
				if(x % a[i - 1])
					mp[x % a[i - 1]] += y;
				mp[a[i - 1]] += x / a[i - 1] * y;
				mp.erase(x);
			}
			else
				break;
		}

	for(auto x : mp)
		t[x.first] += x.second;

	for(int i = n; i >= 0; i--)
		t[i] += t[i + 1];

	for(int i = 1; i <= n; i++)
		cout << t[i] << endl;
}
















