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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <map>
#include <iomanip>
#include <stdio.h>
#include <bitset>

typedef long long LL;

constexpr long long INFLL = 1e18;

using namespace std;

vector<int> a;

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

    a.resize(n + 1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a.at(i);
    }

    int count;
    /*
    vector<int> b(n + 1);
    int min, max;
    for (int i = 1; i <= k; i++)
    {
        b.resize(n + 1, 0);
        for (int m = 1; m <= n; m++)
        {
            min = m - a.at(m);

            max = m + a.at(m);
            if (min < 0)
            {
                min = 0;
            }
            if (max > n)
            {
                max = n;
            }
            for (int l = min; l <= max; l++)
            {
                b.at(l)++;
            }
        }
        a = b;
    }
    */

    vector<int> b_min(n + 1);
    vector<int> b_max(n + 1);
    vector<int> b_count(n + 1);

    for (int i = 1; i <= n; i++)
    {
        b_min.at(i) = i;
        b_max.at(i) = i;
        b_count.at(i) = 1;
    }

    int min, max;
    for (int i = 1; i <= k; i++)
    {
        for (int m = 1; m <= n; m++)
        {
            if (min == 0)
            {
                continue;
            }
            if (max == n)
            {
                continue;
            }
            min = m - a.at(m);
            max = m + a.at(m);
            if (min < 0)
            {
                min = 0;
            }
            if (max > n)
            {
                max = n;
            }

            for (int l = b_min.at(m) - 1; l >= min; l--)
            {
                b_count.at(l)++;
            }
            for (int l = b_max.at(m) + 1; l <= max; l++)
            {
                b_count.at(l)++;
            }
            b_min.at(m) = min;
            b_max.at(m) = max;
        }
        a = b_count;
    }

    for (int i = 1; i <= n; i++)
    {
        cout << a.at(i) << endl;
    }
    return 0;
}