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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i, n) for(ll i=0; i<(ll)n; i++)
#define REPR(i, n) for(ll i=(ll)n-1; i>=0; i--)
#define FOR(i, m, n) for(ll i=(ll)m; i<(ll)n; i++)
#define FORR(i, m, n) for(ll i=(ll)m-1; i>=(ll)n; i--)
#define ALL(x) (x).begin(), (x).end()
#define ASC(v) sort(ALL(v));
#define DESC(v) sort(ALL(v), greater<int>());
#define UNIQ(a) sort(ALL(a));(a).resize(unique(ALL(a)) - a.begin());
#define PRINTD(x, d) cout << fixed << setprecision(d) << x << "\n";
#define DEBUG(x) cout<<#x<<": "<<x<<"\n"
const double PI = acos(-1);
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
bool isprime(int x){ int i; if(x<2)return 0; else if(x==2) return 1; if(x%2==0) return 0; for(i=3; i*i<=x; i+=2) if(x%i==0) return 0; return 1; }
int digsum(int n) { int r=0; while(n>0) { r+=n%10; n/=10; } return r; }
vector<string> split(string s, char p){vector<string>v;stringstream m(s);string b;while(getline(m,b,p))v.push_back(b);return v;}
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
struct nya{ nya(){ cin.tie(0); ios::sync_with_stdio(false); } }nyan;

int main(){
    ll N, K; cin >> N >> K;
    vector<ll> A(N+1);
    FOR(i, 1, N+1) cin >> A[i];
    if (N*K > 1e6 || K*K>N) {
        REP(i, N) cout << N << " ";
        cout << "\n";
        return 0;
    }
    REP(k, K) {
        vector<ll> dp(N+1, 0);
        FOR(n, 1, N+1) {
            FOR(i, n-A[n], n-A[n]+A[n]*2+1) {
                if (i<1 || N<i) continue;
                dp[i]++;
            }
        }
        A = dp;
    }
    FOR(n, 1, N+1) {
        cout << A[n] << " ";
    }
    cout << "\n";
}