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<algorithm>
#include<bitset>
#include<cassert>
#include<cfloat>
#include<climits>
#include<cmath>
#include<deque>
#include<functional>
#include<iomanip>
#include<iostream>
#include<map>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<string>
#include<unordered_map>
#include<unordered_set>
#include<utility>
#include<vector>

using namespace std;

using lint = long long;
using P = pair<int, int>;
using LLP = pair<long long, long long>;

#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())
#define REVERSE(x) reverse((x).begin(), (x).end())
#define TWINS(x) cout << ((x) ? "Yay!" : ":(") << '\n'

constexpr int IINF = (1 << 30) - 1;
constexpr long long LLINF = 1LL << 61;
constexpr double EPS = 1e-10;

constexpr int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
constexpr int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};

template<typename T>
bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<typename T>
bool chmin(T& a, T b){
    if(b < a){
        a = b;
        return true;
    }
    return false;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    int n, k;
    cin >> n >> k;

    vector<int> a(n);
    rep(i, n){
        cin >> a[i];
    }

    vector<int> imos(n + 1, 0);
    rep(i, n){
        ++imos[max(0, i - a[i])];
        --imos[min(n, i + a[i] + 1)];
    }
    rep(i, n){
        imos[i + 1] += imos[i];
    }

    bool flg = true;
    vector<int> dif(n);
    for(int t = 0 ; t < k && flg ; ++t){
        flg = false;
        rep(i, n){
            if(a[i] < n){
                flg = true;
            }
            a[i] = imos[i];
        }
        fill(imos.begin(), imos.end(), 0);
        rep(i, n){
            ++imos[max(0, i - a[i])];
            --imos[min(n, i + a[i] + 1)];
        }
        rep(i, n){
            imos[i + 1] += imos[i];
        }
    }

    rep(i, n){
        if(i) cout << ' ';
        cout << a[i];
    }
    cout << endl;

    cout << flush;

    return 0;
}