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)

// Never let them see you bleed...

#include<bits/stdc++.h>

#define F first
#define S second
#define PB push_back
#define sz(s) int((s).size())
#define bit(n,k) (((n)>>(k))&1)

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;

const int maxn = 1e5 + 10, mod = 1e9 + 7, inf = 1e9 + 10;

ll a[maxn], ans[maxn];

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie();

    int n, q;
    cin >> n >> q;
    for(int i = 1; i <= q; i++){
	cin >> a[i];
    }
    a[0] = n;
    
    map<ll, ll> mp;
    mp[a[q]] = 1;
    
    for(int i = q-1; i >= 0; i--){
	while((mp.rbegin()->F) > a[i]){
	    pll p = *mp.rbegin();
	    mp.erase(p.F);
	    mp[a[i]]+= (p.F / a[i]) * p.S;
	    mp[p.F % a[i]]+= p.S;
	}
    }
    for(pll p : mp){
	ans[p.F]+= p.S;
    }
    for(int i = n-1; i >= 1; i--){
	ans[i]+= ans[i+1];
    }
    for(int i = 1; i <= n; i++){
	cout << ans[i] << "\n";
    }
    return 0;
}