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>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double
#define PI 3.14159265358979323846264338327950288419

#define PII pair<int,int>
#define PLL pair<LL,LL>
#define mp make_pair
#define fi first
#define se second

using namespace std;

int n, m, p;
LL s[100005], u[100005];
PLL t[100005];
LL freq[100005], ans[100005];

void solve(){
	freq[p] = 1;
	FD(i, p, 1){
		if(i == 1) ans[u[1]] += freq[i];
		else if(freq[i] > 0){
			LL temp = u[i];
			while(temp > u[1]){
				int l = 0, r = i;
				while(r - l > 1){
					int mid = (l + r) >> 1;
					if(u[mid] < temp) l = mid;
					else r = mid;
				}
				freq[l] += freq[i] * (temp / u[l]);
				temp %= u[l];
			}
			ans[temp] += freq[i];
		}
	}
}

int main(){
	scanf("%d %d", &n, &m);
	t[0] = mp(n, 0);
	FI(i, 1, m){
		scanf("%lld", &s[i]);
		t[i] = mp(s[i], -i);
	}
	sort(t, t + m + 1);
	int cp = -1;
	FI(i, 0, m){
		if(-t[i].se > cp){
			cp = -t[i].se;
			u[++p] = t[i].fi;
		}
	}
	solve();
	FD(i, n - 1, 1) ans[i] += ans[i + 1];
	FI(i, 1, n) printf("%lld\n", ans[i]);
	return 0;
}