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

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

Examples

Input

5 1
1 0 0 1 0


Output

1 2 2 1 2


Input

5 2
1 0 0 1 0


Output

3 3 4 4 3
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define f(i, x, n) for (int i = x; i < (int)(n); ++i)

int const N = 200000;
int n, k, x[N], y[N];

bool sad(){
	f(i, 0, n)if (x[i] < n)return true;
	return false;
}

void ad(int i, int v) { i = max(i, 0); if (i < n)x[i] += v; }

void up(){
	f(i, 0, n)y[i] = x[i], x[i] = 0;
	f(i, 0, n){
		ad(i - y[i], 1);
		ad(i + y[i] + 1, -1);
	}
	f(i, 1, n)x[i] += x[i - 1];
}

int main(){
	scanf("%d%d", &n, &k);
	f(i, 0, n)scanf("%d", x + i);
	while (k && sad())up(), --k;
	f(i, 0, n)printf("%d%c", x[i], " \n"[i + 1 == n]);
}