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 <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int N = 1e5 + 5;

typedef long long LL;

int n , m , tot;
LL cnt[N] , S[N] , stk[N] , num[N] , x;

int main(void) {
	cin >> n >> m;
	stk[++ tot] = n;
	for(int i = 1;i <= m;++ i) {
		scanf("%lld" , &x);
		while(tot && stk[tot] >= x) tot --;
		stk[++ tot] = x;
	}
	num[tot] = 1;
	int whi;
	for(int i = tot;i >= 1;-- i) {
		if(i == 1) {
			S[stk[i]] += num[i];
			break;
		}
		num[i - 1] += (stk[i] / stk[i - 1]) * num[i];
		stk[i] %= stk[i - 1];
		while(stk[i] >= stk[1]) {
			whi = upper_bound(stk + 1 , stk + i , stk[i]) - stk - 1;
			num[whi] += num[i] * (stk[i] / stk[whi]); stk[i] %= stk[whi];
		}
		S[stk[i]] += num[i];
	}
	for(int i = n;i >= 1;-- i) S[i] += S[i + 1];
	for(int i = 1;i <= n;++ i)printf("%lld\n" , S[i]);
}