Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;

class PL implements Comparable<PL>{
    long x,y;
    PL(long x,long y){
        this.x=x;this.y=y;
    }
    @Override
    public int compareTo(PL o){
        return Long.compare(x,o.x);
    }
    @Override
    public String toString(){
        return "("+x+","+y+")";
    }
}

class Main {
    Main(int n,long[]k){
        int m=k.length;
        Queue<PL> q=new PriorityQueue<PL>(Collections.reverseOrder());
        long f=k[0];
        long[]a=new long[n];
        long c=1;
        for(int i=m-2;i>=0;--i){
            long nc=0;
            while(q.size()>0&&q.peek().x>k[i]){
                PL t=q.poll();
                long x=t.x,y=t.y;
                long qq=x/k[i],r=x%k[i];
                nc+=y*qq;
                if(r>f)
                    q.add(new PL(r,y));
                else if(r>=1)
                    a[(int)(r-1)]+=y;
            }
            long qq=k[i+1]/k[i],r=k[i+1]%k[i];
            nc+=c*qq;
            if(r>f)
                q.add(new PL(r,c));
            else if(r>=1)
                a[(int)(r-1)]+=c;
            c=nc;
        }
        a[(int)(f-1)]+=c;
        while(q.size()>0){
            PL t=q.poll();
            long x=t.x,y=t.y;
            if(x>=1)
                a[(int)(x-1)]+=y;
        }
        for(int i=n-2;i>=0;--i)a[i]+=a[i+1];
        for(long aa:a)System.out.println(aa);
    }
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n=Integer.parseInt(scan.next());
        int q=Integer.parseInt(scan.next());
        int p=1;
        long[] k=new long[q+1];
        k[0]=n;
        for(int i=0;i<q;++i){
            long u=Long.parseLong(scan.next());
            while(p>0&&k[p-1]>=u)p--;
            k[p++]=u;
        }
        new Main(n,Arrays.copyOf(k,p));
    }
}