Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

5 3
6
4
11

Sample Output 1

3
3
3
2
0

After the first operation, the sequence becomes: 1,2,3,4,5,1.

After the second operation, the sequence becomes: 1,2,3,4.

After the third operation, the sequence becomes: 1,2,3,4,1,2,3,4,1,2,3.

In this sequence, integers 1,2,3,4,5 appear 3,3,3,2,0 times, respectively.

Sample Input 2

10 10
9
13
18
8
10
10
9
19
22
27

Sample Output 2

7
4
4
3
3
2
2
2
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

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

typedef long long LL;
const int N = 1e5 + 10;
int n, Q, top; LL q[N], w[N], w0[N];

template <class T> inline void in(T &x) {
	x = 0; int f = 1; char ch = getchar();
	for (; ch<'0' || ch>'9';) {if (ch=='-') f=-1; ch = getchar();}
	for (; ch>='0' && ch<='9';) x = x*10 + ch-'0', ch = getchar();
	x *= f;
}

int main() {
	LL l, x; in(n), in(Q), q[0] = n;
	for (int i = 1; i <= Q; ++i) {
		for (in(x); top >= 0 && q[top] >= x;) --top;
		q[++top] = x;
	}
	w[top] = 1;
	for (int j, i = top; ~i; --i) {
		for (l = q[j = i]; l > q[0];) {
			j = lower_bound(q, q + j, l) - q - 1;
			w[j] += l / q[j] * w[i], l %= q[j];
		}
		w0[l] += w[i];
	}
	for (int i = q[0]; i; --i) w0[i] += w0[i + 1];
	for (int i = 1; i <= n; ++i) printf("%lld\n", w0[i]);
	return 0;
}