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>
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REPLL(i, n) for (ll i = 0; i < (ll)(n); i++)
using namespace std;
template<class T>inline bool chmax(T &a, const T &b){if(a < b){a = b; return 1;}return 0;}
template<class T>inline bool chmin(T &a, const T &b){if(a > b){a = b; return 1;}return 0;}
typedef long long ll;

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll N, K; cin >> N >> K;
  vector<ll> A(N, 0);
  REPLL(i, N){
    cin >> A[i];
  }

  while(K > 0){
    vector<ll> AA(N+2, 0);
    REPLL(i, N){
        if(A[i] != 0){
            ll l = ((i+1)-A[i] < 0 ? 0 : (i+1)-A[i]);
            ll r1 = ((i+1)+1 > N ? N+1 : (i+1)+1);
            ll r2 = ((i+1)+A[i]+1 > N ? N+1 : (i+1)+A[i]+1);
            AA[l]++;
            AA[i+1]--;
            AA[r1]++;
            AA[r2]--;
        }
    }
    for(ll i = 1; i <= N; i++){
        AA[i] = AA[i] + AA[i-1];
        A[i-1] = AA[i]+1;
    }
    if(A[0] == N && A[N-1] == N){
      for(ll i = 0; i < N; i++){
        cout << A[i];
        if(i == N-1){
          cout << endl;
        }else{
          cout << " ";
        }
      }
      return 0;
    }
    K--;
  }

  for(ll i = 0; i < N; i++){
    cout << A[i];
    if(i == N-1){
      cout << endl;
    }else{
      cout << " ";
    }
  }
}