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;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, K;
  cin >> N >> K;
  vector<int> A(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  for (int k = 0; k < K; k++) {
    vector<int> B(N, 0);
    for (int i = 0; i < N; i++) {
      if (A[i] > 0) {
        B[max(0, i - A[i])]++;
        if (i + A[i] + 1 <= N - 1) B[i + A[i] + 1]--;
      }
    }
    bool flag = true;
    for (int i = 0; i < N; i++) {
      if (i + 1 < N) B[i + 1] += B[i];
      if (A[i] == 0) B[i]++;
      if (B[i] != N) flag = false;
    }
    swap(A, B);
    if (flag) break;
  }
  for (auto &v : A) {
    cout << v << (&v == &A.back() ? '\n' : ' ');
  }
  return 0;
}