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;
using ll = long long;
const int maxn = 2e5 + 555;
const int MOD = 1e9 + 7;
int n, k;
int chafeng[maxn];
void add(int l, int r) {
    l = max(1, l);
    r = min(n, r);
    chafeng[l]++;
    chafeng[r + 1]--;
}
bool solve(vector<int> &a) {
    for (int i = 1; i <= n + 1; i++)
        chafeng[i] = 0;
    for (int i = 1; i <= n; i++) {
        add(i - a[i], i + a[i]);
    }
    int now = 0;
    bool ok = true;
    for (int i = 1; i <= n; i++) {
        now += chafeng[i];
        a[i] = now;
        ok &= (a[i] == n);
    }
    return ok;
}
int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    cout << fixed;
    cin >> n >> k;
    vector<int> V(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> V[i];
    for (int i = 1; i <= k; i++) {
        if (solve(V))
            break;
    }
    for (int i = 1; i <= n; i++)
        cout << V[i] << ' ';
    cout << endl;
    return 0;
}