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 <stdio.h>
#pragma GCC optimize ("O2")
#pragma GCC target ("avx2")

#define NMAX 200010

int a[NMAX];
int imos[NMAX];

void reset(int n) {
    for(int i = 0; i < n; i++) imos[i] = 0;
}

int max(int a, int b) {
    if(a > b) return a;
    else return b;
}

int min(int a, int b) {
    if(a > b) return b;
    else return a;
}

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

    for(int itr = 0; itr < k; itr++) {
        int f = 0;
        reset(n+1);
        for(int i = 0; i < n; i++){
            if(a[i] != n) f = 1;
            imos[max(0, i - a[i])]++;
            imos[min(n, i + a[i] + 1)]--;
        }
        int acc = 0;
        for(int i = 0; i < n; i++){
            acc += imos[i];
            a[i] = acc;
        }
        if(!f) break;
    }

    for(int i = 0; i < n; i++) printf("%d\n", a[i]);
    return 0;
}