Problem:
Score : 500 points

Problem Statement

We have N bulbs arranged on a number line, numbered 1 to N from left to right.
Bulb i is at coordinate i.

Each bulb has a non-negative integer parameter called intensity.
When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5.
Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row:

For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i.

Find the intensity of each bulb after the K operations.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq 2 \times 10^5

0 \leq A_i \leq N

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \ldots A_N

Output

Print the intensity A{'}_i of each bulb i after the K operations to Standard Output in the following format:

A{'}_1 A{'}_2 \ldots A{'}_N

Sample Input 1

5 1
1 0 0 1 0

Sample Output 1

1 2 2 1 2

Initially, only Bulb 1 illuminates coordinate 1, so the intensity of Bulb 1 becomes 1 after the operation.
Similarly, the bulbs initially illuminating coordinate 2 are Bulb 1 and 2, so the intensity of Bulb 2 becomes 2.

Sample Input 2

5 2
1 0 0 1 0

Sample Output 2

3 3 4 4 3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define N 200050
#define M 200
#define int long long
#define pb push_back
#define x first
#define y second
#define all(v) v.begin(),v.end()
#define coffee prince
#define sz(a) (int)a.size()
#define ll long long
#define SIZE N
#define MOD mod
using namespace std;
int n,k,ar[N],pr[N];
signed main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
   cin>>n>>k;
   for(int i=0;i<n;i++){
       cin>>ar[i];
   }
   for(int i=0;i<k;i++){
       int fl = 0;
       for(int i=0;i<n;i++){
           if(ar[i]!=n)fl = 1;
       }
       if(!fl)break;
       memset(pr,0,sizeof(pr));
       for(int i=0;i<n;i++){
           int l = max(0ll,i-ar[i]);
           int r = min(n+1,ar[i]+i+1);
           pr[l]++;
           pr[r]--;
       }
       ar[0] = pr[0];
       for(int i=1;i<n;i++){
           pr[i]+= pr[i-1];
           ar[i] = pr[i];
       }
   }
   for(int i=0;i<n;i++)
    cout<<ar[i]<<" ";
  	return 0;
}