Educational Codeforces Round 48 (Rated for Div. 2)——A. Death Note ##

A. Death Note

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You received a notebook which is called Death Note. This notebook has infinite number of pages. A rule is written on the last page (huh) of this notebook. It says: "You have to write names in this notebook during Death Notewith a some strange rule written in it?).

Of course, you decided to follow this rule. When you calmed down, you came up with a strategy how you will write names in the notebook. You have calculated that each page of the notebook can contain exactly always turn the page when it ends, it doesn't matter if it is the last day or not. If after some day the current page still can hold at least one name, during the next day you will continue writing the names from the current page.

Now you are interested in the following question: how many times will you turn the page during each day? You are interested in the number of pages you will turn each day from n.

Input

The first line of the input contains two integers 1≤m≤109) — the number of days you will write names in the notebook and the number of names which can be written on each page of the notebook.

The second line contains i-th day.

Output

Print exactly i-th day.

Examples

input

Copy

3 5
3 7 9

output

Copy

0 2 1 

input

Copy

4 20
10 9 19 2

output

Copy

0 0 1 1 

input

Copy

1 100
99

output

Copy

0 

Note

In the first example pages of the Death Note will look like this [1,1,1,2,2],[2,2,2,2,2],[3,3,3,3,3],[3,3,3,3]. Each number of the array describes during which day name on the

corresponding position will be written. It is easy to see that you should turn the first and the second page during the second day and the third page during the third day.

---

$这个题特别简单，没有什么好说的，就是在这一天写最后写不到一整页的，分到第二天去写，因为翻页是按照写完这一页来看的$

$假设一页可以写4个名字，这一天打算写9个。$

$第一页写完，翻页第一次，还剩5个，$

$第二页写完，翻页第二次，还剩1个，$

$剩余的这一个可以写在第三页上，而为了方便计算，我们把剩余的这一页算在第二天上，于是计算第二天的时候又可以像第一天一样。$

$所以只要计算取余和取整就可以了。$

#include<iostream>
using namespace std;
const int N=2e5+6;
typedef long long ll;
ll a[N],ans[N];int n,m;
int main(){
    ios::sync_with_stdio(false);cin.tie(0);
    cin>>n>>m;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++){
        int x=a[i]%m;a[i+1]+=x;
        ans[i]=a[i]/m;
    }
    for(int i=1;i<=n;i++)cout<<ans[i]<<" ";
}