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 <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
typedef long long ll;
typedef vector<ll> vl;
#define rp(i, f, t) for (long long int i = f; i < t; i++)
#define pr(i, f, t) for (long long int i = t - 1; i >= f; i--)
#define ca(n, a) rp(ca_i, 0, n) cout << a[ca_i] << ((ca_i == n - 1) ? "\n" : " ")
#define ia(n, a) rp(ia_i, 0, n) cin >> a[ia_i]
#define za(n, a) rp(za_i, 0, n) a[za_i] = 0
#define be(a) a.begin(), a.end()
#define ce(a) cout << (a) << endl
int main()
{
    ll n, k;
    cin >> n >> k;
    ll a[n + 1], c[n + 1];
    rp(i, 0, n) cin >> a[i];

    rp(i, 0, k)
    {
        rp(j, 0, n + 1) c[j] = 0;
        rp(j, 0, n)
        {
            if (a[j] >= 1)
            {
                c[j + 1]++;
                c[min(n - 1, j + a[j]) + 1]--;
                c[max(ll(0), j - a[j])]++;
                c[j]--;
            }
        }
        // ca(n, c);
        rp(j, 1, n) c[j] += c[j - 1];
        rp(j, 0, n) a[j] = 1 + c[j];
        bool sig = true;
        rp(j, 0, n) if (a[j] != n)
        {
            sig = false;
            break;
        }
        if (sig)
            break;
    }
    ca(n, a);
}