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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>; 
using pll = pair<long long, long long>;
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007LL;
constexpr long long INF = 1001001001LL;
constexpr long long LINF = 1001001001001001001;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define rept(i, j, n) for(int i=(j); i<(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }


int main(){
     int n, k; cin >> n >> k;
     vector<int> A(n); rep(i, n) cin >> A[i];

     rep(i, min(1000, k)){
         
         vector<int> plus(n), minus(n);
      rep(j, n){
        int s = max(0, j-A[j]), t = min(n-1, j+A[j]);
        plus[s]++; minus[t]++;
      }
      int cnt = 0;
      rep(j, n){
        cnt += plus[j];
        A[j] = cnt;
        cnt -= minus[j];
     }
     }
     rep(i, n)cout << A[i] << " ";
     cout << ln;
}