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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

#define TRACE(x) cerr << #x << " :: " << x << endl
#define _ << " " <<
#define SZ(x) (int)(x).size()
#define ALL(x) (x).begin(),(x).end()
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define RFOR(i,a,b) for (int i=(a);i>=(b);--i)

const int mxN = 1e5+5;
int N, K, A[mxN];
int ft[mxN];

void u(int i, int x) {
    for (; i <= N; i += i&-i) ft[i] += x;
}
void update(int i, int j, int x) {
    u(i,x);
    u(j+1,-x);
}
int query(int i) {
    int r = 0;
    for (; i; i -= i&-i) r += ft[i];
    return r;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> K;
    FOR(i,1,N){
        cin >> A[i];
        update(max(1,i-A[i]),min(N,i+A[i]),1);
    }

    FOR(i,1,K){
        bool brk = 1;
        FOR(j,1,N){
            A[j] = query(j);
            brk &= A[j] == N;
        }
        if (brk) break;
        memset(ft,0,sizeof ft);
        FOR(j,1,N){
            update(max(1,j-A[j]),min(N,j+A[j]),1);
        }
    }
    FOR(i,1,N){
        cout << A[i] << ' ';
    }
    cout << '\n';
}