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>

int main (void){
    long int n,k;
    scanf("%ld %ld",&n,&k);
    long int a[(int)n];
    long int sum=0;
    for(int i=0;i<n;i++)scanf("%ld",&a[i]);

    for(int i = 0;i<k-1;i++){
        long int  b[200005] = {0};
        for(int j=0;j<n;j++){//各電球を見る
            long int min = j-a[j]>0?j-a[j]:0;
            long int max = j+a[j]+1<n?j+a[j]+1:n;
            b[min] ++;//照らし始め
            b[max]--;//終わり
        }
        for(int j=0;j<n;j++){
            sum += b[j];
            a[j] = sum;
        }
        sum = 0;
    }


        long int  b[200005] = {0};
        for(int j=0;j<n;j++){//各電球を見る
            long int min = j-a[j]>0?j-a[j]:0;
            long int max = j+a[j]+1<n?j+a[j]+1:n;
            b[min] ++;//照らし始め
            b[max]--;//終わり
        }
        for(int j=0;j<n;j++){
            sum += b[j];
            a[j] = sum;
        }
        sum = 0;

    for(int i=0;i<n-1;i++)printf("%ld ",a[i]);
    printf("%ld",a[n-1]);


    return 0;
}