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<iostream>
using namespace std;

int main(){
    long long N;
    long long K;
    cin >> N >> K;
    long long A[N+1];
    long long i,j;
    for(i=1; i<=N; i++){
        cin >> A[i];
    }
    long long sum[N+2];
    for(i=1; i<=min((long long)100,K); i++){
        for(j=0; j<=N+1; j++){
            sum[j]=0;
        }
        for(j=1; j<=N; j++){
            if(j-A[j]>=1){
                sum[j-A[j]]++;
            }else{
                sum[1]++;
            }
            if(j+A[j]<=N){
                sum[j+A[j]+1]--;
            }
        }
        for(j=1; j<=N; j++){
            sum[j]=min(sum[j]+sum[j-1],N);
        }
        for(j=1; j<=N; j++){
            if(A[j]<N){
                A[j]=sum[j];
            }
        }
    }
    for(i=1; i<=N-1; i++){
        cout << A[i] << " ";
    }
    cout << A[N] << endl;
    system("pause");
    return 0;
}