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 rep(i, a, b) for(int i = a; i <= b; i++)
#define per(i, a, b) for(int i = a; i >= b; i--)
typedef long long LL;
const int maxn = 1e5 + 5;
const LL mod = 1e9 + 7;

int n, q, len;
LL a[maxn], f[maxn], c[maxn];

void Solve(LL d, LL w) {
	int j = upper_bound(a + 1, a + len + 1, d) - a - 1;
	if(!j) c[1] += w, c[d + 1] -= w;
	else f[j] += d/a[j]*w, Solve(d%a[j], w);
}

int main() {
	cin>>n>>q;
	a[++len] = n;
	while(q--) {
		LL x;
		cin>>x;
		while(a[len] >= x && len) len--;
		a[++len] = x; 
	}
	f[len] = 1;
	per(i, len, 2) {
		f[i - 1] += a[i]/a[i - 1]*f[i];
		Solve(a[i]%a[i - 1], f[i]);
	}
	c[1] += f[1];
	c[a[1] + 1] -= f[1];
	rep(i, 1, n) {
		c[i] += c[i - 1];
	    cout<<c[i]<<endl;
	}
}