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 <bits/stdc++.h>
using namespace std;

const int MAXN = 2e5 + 10;

int N, K;
int A[MAXN];

int sum[MAXN];
int n_count = 0;

void do_operation() {
  int cost = 0;
  for (int i = 1; i <= N; ++i) 
    sum[i] = 0;
  for (int i = 1; i <= N; ++i) {
    sum[max(1, i - A[i])]++;
    sum[min(N + 1, i + A[i] + 1)]--;
  }
  n_count = 0;
  for (int i = 1; i <= N; ++i) {
    cost += sum[i];
    A[i] = cost;
    if (cost == N) ++n_count;
  }
}

void solve() {
  for (int i = 1; i <= N; ++i)
    if (A[i] == N)
      n_count += 1;
  for (int k = 1; k <= K; ++k) {
    if (n_count == N) break;
    do_operation();
  }
}

int main() {
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; ++i) scanf("%d", &A[i]);
  solve();
  for (int i = 1; i <= N; ++i) printf("%d ", A[i]);
  printf("\n");
  return 0;
}