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: Runtime Error)

#include <bits/stdc++.h>

#define taskname ""
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define for0(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex <ld> cd;
typedef vector <cd> vcd;
typedef vector <int> vi;

template<class T> using v2d = vector <vector <T> >;
template<class T> bool uin(T &a, T b)
{
    return a > b ? (a = b, true) : false;
}
template<class T> bool uax(T &a, T b)
{
    return a < b ? (a = b, true) : false;
}

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

const int maxN = 1e5 + 10;

int n, k;
ll a[maxN], b[maxN];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> k;
    for1(i, n)
    {
        cin >> a[i];
    }
    uin(k, 20000000 / n);
    while (k--)
    {
        fill(b + 1, b + n + 1, 0);
        for1(i, n)
        {
            int l = max(i - a[i], 1LL), r = min(i + a[i], (ll)n);
            ++b[l];
            if (r < n)
            {
                --b[r + 1];
            }
        }
        for1(i, n)
        {
            b[i] += b[i - 1];
            a[i] = b[i];
        }
    }
    for1(i, n)
    {
        cout << a[i] << ' ';
    }
    return 0;
}