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.

Examples

Input

5 3
6
4
11


Output

3
3
3
2
0


Input

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


Output

7
4
4
3
3
2
2
2
0
0
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;int tp=0;long long dp[100005],sta[100005],pre[100005];
inline void dac(long long len,long long val) {
	int pos=upper_bound(sta+1,sta+tp+1,len)-sta-1;
	if (!pos) return pre[1]+=val,pre[len+1]-=val,void();
	dp[pos]+=val*(len/sta[pos]);return dac(len%sta[pos],val);
}int main() {int n,q;long long x;scanf("%d%d",&n,&q),sta[tp=1]=n;
	for (int i=1;i<=q;++i) {scanf("%lld",&x);
		while (tp&&sta[tp]>=x) --tp;sta[++tp]=x;
	}dp[tp]=1;for (int i=tp-1;i;--i) {
		dp[i]+=dp[i+1]*(sta[i+1]/sta[i]),dac(sta[i+1]%sta[i],dp[i+1]);
	}pre[1]+=dp[1],pre[sta[1]+1]-=dp[1];
	for (int i=1;i<=n;++i) pre[i]+=pre[i-1],printf("%lld\n",pre[i]);return 0;
}
