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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
    int N, K;
    cin >> N >> K;
    vector<int> A(N);
    for (int i = 0; i < N; i++)
    {
        cin >> A[i];
    }
    
    if (K >= N)
    {
        for (int i = 0; i < N; i++)
        {
            cout << N;
            if (i != N - 1)
            {
                cout << " ";
            } else
            {
                cout << endl;
            }
    
        }
        return 0;
    }
    
    for (int k = 1 ; k <= K; k++) {
        vector<int> B(N);
        for (int i = 0; i < N; i++)
        {
            int x = ceil(i + 1 - A[i] - 0.5);
            int y = floor(i + 1 + A[i] + 0.5);
            cerr << "i=" << i << endl;
            cerr << "x=" << x << ",y=" << y << endl;
            for (int j = max(x, 1); j <= min(y, N); j++)
            {
                B[j - 1]++;
            }            
        
        }
        cerr << "k=" << k << endl;
        cerr << "B[0]=" << B[0] << ",B[1]=" << B[1] << endl;
        A = B;
        if (B[0] == N && B[N - 1] == N)
        {
            break;
        }
                        
    }
    
    for (int i = 0; i < N; i++)
    {
        cout << A[i];
        if (i != N - 1)
        {
            cout << " ";
        } else
        {
            cout << endl;
        }
        
        
    }
    
  return 0;
}