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 <stdbool.h>
#include <stdio.h>
int a[200000], tmpa[200000], light[200010] = {0};
typedef long long ll;
int main() {
	int n, k;
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}

	for (int i = 0; i < n; i++) {
		int r, l;
		r = i - a[i];
		l = i + a[i];
		light[(r < 0 ? 0 : r)]++;
		light[(l + 1 > n ? n + 1 : l + 1)]--;
	}
	// for (int i = 0; i <= n; i++) {
	// 	printf("%d ", light[i]);
	// }
	// printf("\n");
	for (int i = 0; i < k; i++) {
		int sum = 0;
		for (int j = 0; j < n; j++) {
			tmpa[j] = a[j];
			sum += light[j];
			a[j] = sum;
		}
		for (int j = 0; j < n; j++) {
			int r = j - tmpa[j];
			int l = j + tmpa[j];
			light[(r < 0 ? 0 : r)]--;
			light[(l + 1 > n ? n + 1 : l + 1)]++;
			r = j - a[j];
			l = j + a[j];
			light[(r < 0 ? 0 : r)]++;
			light[(l + 1 > n ? n + 1 : l + 1)]--;
		}
		bool isSame = true;
		for (int j = 0; j < n - 1; j++) {
			if (a[j] != a[j + 1]) {
				isSame = false;
				break;
			}
		}
		if (isSame && a[0] == n) {
			break;
		}
		// for (int j = 0; j < n; j++) {
		// 	printf("%d", a[j]);
		// 	printf(j == n - 1 ? "\n" : " ");
		// }
	}
	for (int j = 0; j < n; j++) {
		printf("%d", a[j]);
		printf(j == n - 1 ? "\n" : " ");
	}
	return 0;
}