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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, x, y) for (int i = x; i <= y; i++)
using namespace std;

typedef long long ll;
const int N = 1e5 + 10;
ll n, Q, m, q[N], x, f[N], ans[N];

void solve(ll x, ll v) {
    int k = upper_bound(q + 1, q + m + 1, x) - q - 1;
    if (!k) {
        ans[1] += v, ans[x + 1] -= v;  // 提前累计
        return;
    }
    f[k] += v * (x / q[k]), solve(x % q[k], v);
}

int main() {
    cin >> n >> Q;
    q[++m] = n;
    rep(i, 1, Q) {
        scanf("%lld", &x);
        while (m && q[m] >= x) --m;
        q[++m] = x;
    }

    f[m] = 1;
    for (int i = m; i >= 2; i--)
        f[i - 1] += f[i] * (q[i] / q[i - 1]), solve(q[i] % q[i - 1], f[i]);

    ans[1] += f[1], ans[q[1] + 1] -= f[1];
    rep(i, 1, n) ans[i] += ans[i - 1], printf("%lld\n", ans[i]);
    return 0;
}