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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define X first
#define Y second
#define MP make_pair

using namespace std;

typedef long long ll;

const int N = 1e5 + 1;

int n, q;
ll t[N], add[N], sq[N], sz;

int main () {
	scanf("%d %d\n", &n, &q);
	sq[0] = n;
	for(;q;q--){
		ll x;
		scanf("%lld\n", &x);
		while(sz >= 0 && sq[sz] > x){
			sz--;	
		}
		sq[++sz] = x;
	}
	for(int j = sz;j >= 0;j--){
		if(j == sz)
			t[j] = 1;
		for(int idx = j - 1;idx >= 0;idx--){
			int tl = 0, tr = idx;
			bool can = 0;	
			while(tl <= tr){
				int tm = (tl + tr) / 2;
				if(sq[tm] <= sq[j])
					tl = tm + 1, idx = tm, can = 1;	
				else
					tr = tm - 1;
			}
			if(!can)
				break;
			t[idx] += t[j] * (sq[j] / sq[idx]);
			sq[j] %= sq[idx];
		}	
		add[sq[j]] += t[j];
	}
	for(int i = n - 1;i >= 1;i--)
		add[i] += add[i + 1];
	for(int i = 1;i <= n;i++)
		printf("%lld\n", add[i]);
}