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 <bits/stdc++.h>
#include <math.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep_s(i, s, n) for (int i = (int)(s); i < (int)(n); i++)
using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;

#define MOD 1000000007

int main()
{
  int n, k;
  cin >> n >> k;

  vi a(n);
  vector<vector<int>> light_a(n);

  rep(i, n)
  {
    cin >> a[i];
    int start = i - a[i];
    if (start < 0)
    {
      start = 0;
    }
    for (int k = start; k <= i + a[i] && k < n; k++)
    {
      light_a[k].push_back(i);
    }
  }

  rep(_k, k)
  {
    vi table(n + 1, 0);
    rep(i, n)
    {
      table[max(i - a[i], 0)]++;
      table[min(i + a[i] + 1, n)]--;
    }
    rep(i, n)
    {
      if (i > 0)
      {
        table[i] += table[i - 1];
      }
    }

    table.pop_back();
    if (table == a)
    {
      break;
    }

    a = table;
  }

  rep(i, n)
  {
    cout << a[i] << endl;
  }
}