Problem:
Score : 2000 points

Problem Statement

You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.

You can apply the following operation to this permutation, any number of times (possibly zero):

Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1.
Then, swap the values of P_i and P_j.

Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

2≦N≦500,000

1≦K≦N-1

P is a permutation of the set {1, 2, ..., N}.

Input

The input is given from Standard Input in the following format:

N K
P_1 P_2 ... P_N

Output

Print the lexicographically smallest permutation that can be obtained.

Sample Input 1

4 2
4 2 3 1

Sample Output 1

2
1
4
3

One possible way to obtain the lexicographically smallest permutation is shown below:

4 2 3 1

4 1 3 2

3 1 4 2

2 1 4 3

Sample Input 2

5 1
5 4 3 2 1

Sample Output 2

1
2
3
4
5

Sample Input 3

8 3
4 5 7 8 3 1 2 6

Sample Output 3

1
2
6
7
5
3
4
8

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define maxn 100010
#define ll long long
#define db double
#define vi vector<int>
#define pb push_back
#define mod 1000000007
int c[4005][4005];
int a[4005];
const int mxs = 4005;
void read(int &a) {
	char ch = getchar();
	while (ch > '9' || ch < '0') ch = getchar();
	while (ch >= '0' && ch <= '9') a = a * 10 + ch - '0', ch = getchar();
}
int main() {
	for (int i = 0; i < 4005; i++) {
		for (int j = 0; j <= i; j++) {
			if (j == 0 || j == i) c[i][j] = 1;
			else {
				c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
			}
		}
	}
	int test = 0;
	int n;
	if (test) n = 200000;
	else
		read(n);
	const int pls = 2001;
	ll ans = 0;
	for (int i = 1; i <= n; i++) {
		int x, b;
		if (test) b = 1900 + rand() % 100, x = 1900 + rand() % 100;
		else read(x), read(b);
		x += b;
		double nans = 0;
		for (int j = pls - b; j <= min(mxs - 1, x + pls - b); j++) {
			ll q = 1ll * a[j] * c[x][b + j - pls];
			nans = (nans + q), ans = (ans + q);
		}
		ans -= (ll)(nans / mod) * mod;
		ans %= mod;
		if (ans < 0) ans += mod;
	//	cout << ans << ' ' << nans << endl;
		for (int j = b + pls - x; j <= b + pls; j++) {
			a[j] += c[x][b + pls - j];
			a[j] = (a[j] >= mod ? a[j] - mod : a[j]);
		}
	}
	cout << ans << endl;
	return 0;
}