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)

#include <bits/stdc++.h>
using namespace std;

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

    vector<int> inp(n + 1,0);
    for(int i = 1 ; i <= n ; i++){
        //cin >> inp.at(i);
        inp.at(i) = 0;
    }

    vector<int> data(n + 2,0);
    
    if(k >= 2000){
        k = 2000;
    }

    for(int j = 0 ; j < k ; j++){
        for(int i = 1 ; i <= n ; i++){
            int left = max(i - inp.at(i), 1);
            int right = min(i + inp.at(i) + 1, n + 1);
            data.at(left)++;
            data.at(right)--;
        }
        int temp = 0;
        for(int i = 1 ; i <= n ; i++){
            temp += data.at(i);
            data.at(i) = 0;
            inp.at(i) = temp;
        }

        
    }

    //確認用
    for(int i = 1 ; i <= n ; i++){
        cout << inp.at(i);
        if(i != n){
            cout << " ";
        }
    }
    //cout << " " << j+1 << endl;
    cout << endl;
}