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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX = (1e5)+1;

int n, m;
vector<ll> a, need;
priority_queue<pair<ll, ll> > q;
ll f[MAX];
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	a.push_back(n);
	for(int i = 0; i < m; i++) {
		ll x; cin >> x;
		while(!a.empty() && a.back() >= x) a.pop_back();
		a.push_back(x);
	}
	need.resize(a.size());
	need.back() = 1;
	for(int i = a.size()-1; i >= 0; i--) {
		while(!q.empty() && q.top().first >= a[i]) {
			need[i] += (q.top().first / a[i]) * q.top().second;
			if(q.top().first % a[i]) q.emplace(q.top().first % a[i], q.top().second);
			q.pop();
		}
		if(i > 0) {
			if(a[i] % a[i-1]) q.emplace(a[i] % a[i-1], need[i]);
			need[i-1] += (a[i] / a[i-1]) * need[i];
		} else {
			f[a[0]] += need[i];
			while(!q.empty()) {
				assert(q.top().first <= a[0]);
				f[q.top().first] += q.top().second;
				q.pop();
			}
		}
	}
	for(int i = n-1; i >= 1; i--) f[i] += f[i+1];
	for(int i = 1; i <= n; i++) cout << f[i] << '\n';
}