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: Wrong Answer)

// -----------------------------------
// author   : MatsuTaku
// country  : Japan
// created  : 06/13/20 21:03:09
// -----------------------------------

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int n,k; cin>>n>>k;
  vector<int> A(n); for (auto& a:A) cin>>a;
  vector<int> ans(n);
  for (int i = 0; i < 10; i++) {
    ans.assign(n, 0);
    for (int i = 0; i < n; i++) {
      ans[max(0, i-A[i])]++;
      if (i+A[i]+1 < n)
        ans[i+A[i]+1]--;
    }
    for (int i = 1; i < n; i++)
      ans[i] += ans[i-1];
    bool skip = true;
    for (int i = 0; i < n; i++) {
      if (ans[i] < n) {
        skip = false;
        break;
      }
    }
    if (skip)
      break;
    A = ans;
  }
  for (auto a:ans)
    cout << a << " ";
  cout << endl;

  return 0;
}