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 <cstring>
#include <cstdlib>
#include <algorithm>
#define LL long long
using namespace std;
const LL Maxn = 100010;
LL sta[Maxn], tp, n, m;
LL s[Maxn], t[Maxn];
LL bs(LL r, LL x) {
	LL l = 1, ret = 0;
	while(l <= r){
		LL mid = l+r>>1;
		if(sta[mid] <= x){ ret = mid; l = mid+1; }
		else r = mid-1;
	}
	return ret;
}
int main() {
	LL i, j, k;
	scanf("%lld%lld", &n, &m);
	sta[tp=1] = n;
	for(i = 1; i <= m; i++){
		LL x;
		scanf("%lld", &x);
		while(tp && sta[tp] >= x) tp--;
		sta[++tp] = x;
	}
	t[tp] = 1;
	for(i = tp; i >= 1; i--){
		k = sta[i]; LL x = bs(i-1, k);
		while(x){
			t[x] += k/sta[x]*t[i];
			k %= sta[x];
			x = bs(x-1, k);
		}
		s[1] += t[i]; s[k+1] -= t[i];
	}
	for(i = 1; i <= n; i++) printf("%lld\n", s[i] += s[i-1]);
	return 0;
}