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 (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

// 一つづつ見ていくパターン
// 制限時間オーバーのため、要改善


#include <stdio.h>

int main(){
    int n, k;
    int power[200000];
    int temp[200000];

    scanf("%d %d", &n, &k);

    for (int i = 0; i < n; i++){
        scanf("%d", &power[i]);
    }


    int total = 0;
    int preTotal = 0;

    for (int i = 0; i < k; i++) {
        for (int i = 0; i < n; i++){
            temp[i] = 0;
        }

        for (int j = 0; j < n; j++) {
            int s, e;
            if ((j - power[j]) < 0) {
                s = 0;
            } else {
                s = j - power[j];
            }

            if ((j + power[j]) > (n - 1)) {
                e = n - 1;
            } else {
                e = j + power[j];
            }

            for (int l = s; l <= e; l++) {
                temp[l] += 1;
            }

        }

        total = 0;
        for (int j = 0; j < n; j++) {
            power[j] = temp[j];
            total += power[j];
        }

        if (preTotal == total) {
            break;
        }
        preTotal = total;
    }

    for (int i = 0; i < n; i++) {
        if (i != 0) {
            printf(" ");
        }

        printf("%d", power[i]);
    }
}