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;
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define dd(x) cout << #x << "=" << x << ","
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define per(i,a,b) for(int i=(b-1);i>=a;--i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define endl "\n"
#define lowbit(x) x&(-x)
const int N = 20000;
const int M = 1e9+7;

int a[110][200010];
int b[110][200010];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    for (int i=1; i<=n; i++){
    	cin>>a[0][i];
	}
	for (int i=0; i<k; i++){
		for (int j=1; j<=n; j++){
			b[i+1][max(1,j-a[i][j])]++;
			b[i+1][min(n+1,j+a[i][j]+1)]--;
		}
		int t=0,f=1;
		for (int j=1; j<=n; j++){
			t+=b[i+1][j];
			a[i+1][j]=t;
			if (a[i+1][j]!=n) f=0;
		}
		if (f){
			for (int j=1; j<=n; j++)
				cout<<a[i+1][j]<<' ';
			return 0;
		}
	}
	for (int j=1; j<=n; j++)
		cout<<a[k][j]<<' ';
    return 0;
}