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: Wrong Answer)

#include <stdio.h>
 
int main()
{
  int n,k;
  scanf("%d%d",&n,&k);
  int a[n];
  int i;
  i = 0;
  while (i < n)
  {
    scanf("%d",&a[i]);
    i++;
  }
  i = 0;
  /*
  while (i < n)
  {
    printf("%d",a[i]);
    i++;
  }
  */
  int j;
  int m;
  int cnt[n];
  int index;
  index = 0;
  i = 0;
  if (k >= n/2)
  {
    while (i < n)
    {
      printf("%d ",n);
      i++;
    }
    return (0);
  }
  while (i < n)
  {
    cnt[i] = 0;
    i++;
  }
  i = 0;
  int all;
  all = 0;
  while (i < k)
  {
    j = 0;
    while (j < n)
    {
      m = j-a[j];
      if (m < 0)
      {
        m = 0;
      }
      while (m <= j+a[j])
      {
        if (a[j] > j && a[j] > n-j)
        {
        	all++;
        	break;
        }
        //printf("m:%d\n",m);
        //printf("a[m]:%d\n",a[m]);
        cnt[m]++;
        m++;
      }
      j++;
    }
    index = 0;
    while (index < n)
    {
      a[index] = cnt[index] + all;
      cnt[index] = 0;
      index++;
    }
    all = 0;
    i++;
  }
  i = 0;
  while (i < n)
  {
    printf("%d ",a[i]);
    i++;
  }
}