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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef  long long  ll;
const int maxn = 2e5 + 5;
int n, q;
ll a, t[maxn], ans[maxn];
vector<ll> v;
int main() {
    scanf("%d %d",&n,&q);
    v.push_back(n);//!!!
    for (int i=0; i<q; i++) {
        scanf("%lld",&a);
        while (v.size() && v.back() >= a)
            v.pop_back();
        v.push_back(a);
    }
    if ( q== 0) {
        for (int i=0; i<n; i++) puts("1");
        return 0;
    }
    //for (int i=0; i<v.size(); i++) cout<<v[i]<<"  ";
    t[v.size() - 1] = 1;
    for (int i=v.size()-1; i>=0; i--) {
        ll k = v[i];
        //cout<<endl<<i<<" : \n";
        while (k > v[0]) {
            int j = lower_bound(v.begin(), v.end(), k) - (v.begin() + 1);
            t[j] += k / v[j] * t[i];
            //cout<<k<<" "<<j<<" : "<<v[j]<<" "<<t[j]<<" "<<t[i]<<endl;
            k %= v[j];
        }
        ans[k] += t[i];
    }
    for (int i=n-1; i>=1; i--) ans[i] += ans[i+1];
    for (int i=1; i<=n; i++)
        printf("%lld\n", ans[i]);
    return 0;
}