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>
#define maxn 100086

using namespace std;

typedef long long ll;

int n, m;

ll x, sk[maxn], top;
ll f[maxn], a[maxn];

void solve(ll d, ll w){
	int i = upper_bound(sk + 1, sk + 1 + top, d) - (sk + 1);
	if(!i) a[1] += w, a[d + 1] -= w;
	else f[i] += d / sk[i] * w, solve(d % sk[i], w);	
}

int main(){
	scanf("%d%d", &n, &m);
	sk[++top] = n;
	while(m--){
		scanf("%lld", &x);
		while(top && sk[top] >= x) --top;
		sk[++top] = x;
	}
	f[top] = 1;
	for(int i = top - 1;i;i--){
		f[i] += sk[i + 1] / sk[i] * f[i + 1];
		solve(sk[i + 1] % sk[i], f[i + 1]);
	}
	a[1] += f[1], a[sk[1] + 1] -= f[1];
	for(int i = 1;i <= n;i++) printf("%lld\n", a[i] += a[i - 1]);
}