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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

int *t(int n)
{
  int i;
  int *table = calloc(n,sizeof(int));
  for(i=0;i<n;i++)
  {
    table[i] = 0;
  }
  return table;
}

void solve(int n,int k,int *a)
{
  int i,j,l;
  for(i=0;i<k;i++)
  {
    int *table = t(n+1);
    for(j=0;j<n;j++)
    {
      if(0<j-a[j])  
        table[j-a[j]]++;
      else  
        table[0]++;

      if(n-1<j+a[j])  
        table[n-1+1]--;
      else  
        table[j+a[j]+1]--;
    }
    for(l=0;l<n;l++)
    {
      if(l==0)
      {
        a[l] = table[l];
        continue;
      }
      a[l] = a[l-1] + table[l];
    }
    free(table);
  }
  for(i=0;i<n;i++)
    printf("%d ",a[i]);
}

int main()
{
  int n,k;
  scanf("%d %d",&n,&k);
  int *a = calloc(n,sizeof(int));
  int i;
  for(i=0;i<n;i++)
    scanf("%d",&a[i]);
  solve(n,k,a);
  free(a);
  return 0;
}