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<cstdio>
#include<algorithm>
using namespace std;

typedef long long ll;
const int N = 100005;

int n, m, top;
ll a[N], stk[N], f[N], s[N];

// [1, i] repeat j times. 

void work(ll i, ll j)
{
	if(i <= a[1])
	{
		s[1] += j;
		s[i + 1] -= j;
		return;
	}
	int p = lower_bound(a + 1, a + 1 + m, i) - a - 1;
	f[p] += j * (i / a[p]);
	work(i % a[p], j);
}

int main()
{
	scanf("%d %d", &n, &m);
	stk[top = 1] = n;
	for(int i = 1; i <= m; ++i)
	{
		ll x;
		scanf("%lld", &x);
		for(; top && stk[top] >= x; --top);
		stk[++top] = x;
	}
	m = top;
	copy(stk, stk + m + 1, a);
	f[m] = 1;
	a[0] = 4e18;
	for(int i = m; i; --i)
		if(f[i])
			work(a[i], f[i]);
	for(int i = 1; i <= n; ++i)
	{
		s[i] += s[i - 1];
		printf("%lld\n", s[i]);
	}
	return 0;
}
