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 $\mathrm{nn consecutive\; days.\; During\; the ii-th\; day\; you\; have\; to\; write\; exactly aiai names.".\; You\; got\; scared\; (of\; course\; you\; got\; scared,\; who\; wouldn\text{'}t\; get\; scared\; if\; he\; just\; receive\; a\; notebook\; which\; is\; named Death\; Notewith\; a\; some\; strange\; rule\; written\; in\; it?).}$

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 $\mathrm{mm names.\; You\; will\; start\; writing\; names\; from\; the\; first\; page.\; You\; will\; write\; names\; on\; the\; current\; page\; as\; long\; as\; the\; limit\; on\; the\; number\; of\; names\; on\; this\; page\; is\; not\; exceeded.\; When\; the\; current\; page\; is\; over,\; you\; turn\; the\; page.\; Note\; that\; you always turn\; the\; page\; when\; it\; ends,\; it\; doesn\text{'}t\; matter\; if\; it\; is\; the\; last\; day\; or\; not.\; If\; after\; some\; day\; the\; current\; page\; still\; can\; hold\; at\; least\; one\; name,\; during\; the\; next\; day\; you\; will\; continue\; writing\; the\; names\; from\; the\; current\; page.}$

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 $\mathrm{11 to nn.}$

Input

The first line of the input contains two integers $\mathrm{nn, mm (1\le n\le 2\cdot 1051\le n\le 2\cdot 105, 1\le m\le 1091\le m\le 109)\; —\; the\; number\; of\; days\; you\; will\; write\; names\; in\; the\; notebook\; and\; the\; number\; of\; names\; which\; can\; be\; written\; on\; each\; page\; of\; the\; notebook.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109),\; where aiai means\; the\; number\; of\; names\; you\; will\; write\; in\; the\; notebook\; during\; the ii-th\; day.}$

Output

Print exactly $\mathrm{nn integers t1,t2,\dots ,tnt1,t2,\dots ,tn,\; where titi is\; the\; number\; of\; times\; you\; will\; turn\; the\; page\; during\; the ii-th\; day.}$

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 $[1,1,1,2,2],[2,2,2,2,2],[3,3,3,3,3],[3,3,3,3][1,1,1,2,2],[2,2,2,2,2],[3,3,3,3,3],[3,3,3,3].\; Each\; number\; of\; the\; array\; describes\; during\; which\; day\; name\; on\; the$

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

---

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

$假设一页可以写4个名字，这一天打算写9个。$

$第一页写完，翻页第一次，还剩5个，$

$第二页写完，翻页第二次，还剩1个，$

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

$所以只要计算取余和取整就可以了。$

#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]<<" ";
}