Garbage Disposal

Description

Enough is enough. Too many times it happened that Vasya forgot to dispose of garbage and his apartment stank afterwards. Now he wants to create a garbage disposal plan and stick to it.

For each of next nn days Vasya knows aiai — number of units of garbage he will produce on the ii-th day. Each unit of garbage must be disposed of either on the day it was produced or on the next day. Vasya disposes of garbage by putting it inside a bag and dropping the bag into a garbage container. Each bag can contain up to kk units of garbage. It is allowed to compose and drop multiple bags into a garbage container in a single day.

Being economical, Vasya wants to use as few bags as possible. You are to compute the minimum number of bags Vasya needs to dispose of all of his garbage for the given nn days. No garbage should be left after the nn-th day.

Input

The first line of the input contains two integers nn and kk (1≤n≤2⋅105,1≤k≤1091≤n≤2⋅105,1≤k≤109) — number of days to consider and bag's capacity. The second line contains nn space separated integers aiai (0≤ai≤1090≤ai≤109) — the number of units of garbage produced on the ii-th day.

Output

Output a single integer — the minimum number of bags Vasya needs to dispose of all garbage. Each unit of garbage should be disposed on the day it was produced or on the next day. No garbage can be left after the nn-th day. In a day it is allowed to compose and drop multiple bags.

Sample Input

Input

3 2
3 2 1

Output

3

Input

5 1
1000000000 1000000000 1000000000 1000000000 1000000000

Output

5000000000

Input

3 2
1 0 1

Output

2

Input

4 4
2 8 4 1

Output

4

题解：

简单题，但是做错了。。。。每天要解决今天和昨天的垃圾，今天的可以留给明天，最后一天必须全部干完，一次解决k，问需要多少次？

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 200007
#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
using namespace std;
typedef long long ll;

int a[200007],c[200007];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=0;i<n;i++)
        cin>>a[i];
    ll sum=0;
    sum+=a[0]/k;
    int s=a[0]%k;
    for(int i=1;i<n;i++)
    {
        if(s+a[i]>=k)
        {
            sum+=(s+a[i])/k;
            s=(s+a[i])%k;
        }
        else if(s==0&&a[i]==0)
            continue;
        else if(s==0&&a[i]<k)
            s=a[i];
        else
        {
            sum++;
            s=0;
        }
        //cout<<sum<<" "<<s<<endl;
    }
    if(s>0)
        sum++;
    cout<<sum<<endl;
    return 0;
}