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;
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define whole(x) (x).begin(),(x).end()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
using P = pair<int, int>;
#define debug(var) cerr << "[" << #var << "] " << var << endl
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
const ll mod = 1000000007;
const int dx[] = {-1,0,1,0};
const int dy[] = {0,-1,0,1};


int main(){
    ll n, k;
    cin >> n >> k;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    vector<ll> ans(n);
    rep(i, n) ans[i] = a[i];
    rep(i, k) {
        bool flg = true;
        vector<ll> pre(n);
        swap(pre, ans);
        vector<ll> x(n+1);
        rep(j, n) {
            ll l = j - pre[j];
            chmax(l, 0LL);
            ll r = j + pre[j] + 1;
            chmin(r, n);
            x[l] += 1LL;
            x[r] -= 1LL;
        }
        rep(j, n) {
            if (j==0) ans[j] = x[j];
            else {
                ans[j] = ans[j-1] + x[j];
            }
            if (ans[j]<n) flg = false;
        }
        if (flg) break;
        
    }
    
    rep(i, n) cout << ans[i] << " ";
    cout << endl;
    return 0;
}