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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
#define ll long long
#define ull unsigned long long
#define db long double
#define pb push_back
#define ppb pop_back
#define F first
#define S second
#define mp make_pair
#define all(x) (x).begin(), (x).end()

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef array <ll, 2> pii;
typedef vector <int> vi;

typedef tree <pii, null_type, less <pii>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

const int N = 2e5 + 123;
int n, k, a[N], p[N];

int main() {
	ios_base::sync_with_stdio(false), cin.tie(NULL);
	#ifdef LOCAL
		freopen("input.txt", "r", stdin);
	#endif
	
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	int cnt = 0;
	while (k--) {
		memset(& p, 0, sizeof(p));
		for (int i = 1; i <= n; i++) {
			int l = max(1, i - a[i]);
			int r = min(n, i + a[i]);
			p[l]++;
			p[r + 1]--;
		}
		int mn = N;
		for (int i = 1; i <= n; i++) {
			a[i] = a[i - 1] + p[i];
			mn = min(mn, a[i]);
		}
		cnt++;
		if (mn == n) {
			break;
		}
	}
	for (int i = 1; i <= n; i++) {
		cout << a[i] << ' ';
	}
}