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>
#include <unistd.h>
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))

int main(void)
{
    char f = 0, f2, ff; int tmp1, tmp2;
    int i, ii, p;
    int n, k, l[2][200005];
    scanf("%d%d", &n, &k);
    for (i = 0; i < n; i++)
    {
        scanf("%d", l[0] + i);
    }

    f = 0;
    for (p = 0; p < k * 2 / 3; p++)
    {
        for (i = 0; i < n; i++)
        {
            tmp1 = max(i - l[f][i], 0);
            tmp2 = min(i + l[f][i], n - 1);
            f2 = f * (-1) + 1;
            for (ii = tmp1; ii <= tmp2; ii++)
                l[f2][ii]++; 
    
            l[f][i] = 0;
            
        }
        f = f2;
        ff = 0;
        
        
        for (i = 0; i < n; i++)
        {
            if (l[f][i] < n)
            {
                ff = 1;
                break;
            }
        }
        if (ff == 0)
        {
            for (i = 0; i < n - 1; i++)
                printf("%d ", n);
            printf("%d", n);
            return (0);
        }
        
    }
    
    
    
    
    
    
    
    for (p = p; p < k; p++)
    {
        for (i = 0; i < n; i++)
        {
            tmp1 = max(i - l[f][i], 0);
            tmp2 = min(i + l[f][i], n - 1);
            f2 = f * (-1) + 1;
            for (ii = 0; ii <tmp1; ii++)
                l[f2][ii]--; 
            for (ii = tmp2 + 1; ii < n - 1; ii++)
                l[f2][ii]--;
            l[f][i] = n;
            
        }
        f = f2;
        ff = 0;
        
        
        for (i = 0; i < n; i++)
        {
            if (l[f][i] < n)
            {
                ff = 1;
                break;
            }
        }
        if (ff == 0)
        {
            for (i = 0; i < n - 1; i++)
                printf("%d ", n);
            printf("%d", n);
            return (0);
        }
        
    }
    
    
    
    for (i = 0; i < n - 1; i++)
    {
        printf("%d ", l[f][i]);
    }
    printf("%d", l[f][n - 1]);
    return (0);
}