VK Cup 2016 - Round 1 B. Bear and Forgotten Tree 3

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 through n.

Limak is a little polar bear and Radewoosh is his evil enemy. Limak once had a tree but Radewoosh stolen it. Bear is very sad now because he doesn't remember much about the tree — he can tell you only three values n, d and h:

• The tree had exactly n vertices.
• The tree had diameter d. In other words, d was the biggest distance between two vertices.
• Limak also remembers that he once rooted the tree in vertex 1 and after that its height was h. In other words, h was the biggest distance between vertex 1 and some other vertex.

The distance between two vertices of the tree is the number of edges on the simple path between them.

Help Limak to restore his tree. Check whether there exists a tree satisfying the given conditions. Find any such tree and print its edges in any order. It's also possible that Limak made a mistake and there is no suitable tree – in this case print "-1".

Input

The first line contains three integers n, d and h (2 ≤ n ≤ 100 000, 1 ≤ h ≤ d ≤ n - 1) — the number of vertices, diameter, and height after rooting in vertex 1, respectively.

Output

If there is no tree matching what Limak remembers, print the only line with "-1" (without the quotes).

Otherwise, describe any tree matching Limak's description. Print n - 1 lines, each with two space-separated integers – indices of vertices connected by an edge. If there are many valid trees, print any of them. You can print edges in any order.

Examples

Input

5 3 2

Output

1 2
1 3
3 4
3 5

Input

8 5 2

Output

-1

Input

8 4 2

Output

4 8
5 7
2 3
8 1
2 1
5 6
1 5

Note

Below you can see trees printed to the output in the first sample and the third sample.

   好不容易过了第一批测试数据后在比赛快结束时被hack了。。。。我还是TOO SIMPLE。
   细节有好多，-1的情况有两种，2*h < d 或 h=d=1 且 n ！= 2,比赛时就是没注意到第二种情况。
  当h != d 时按点编号顺序构造出最长深度后把d-h剩下的长度补到1点上，补满后其他多余点继续插到1号点上。
  当h=d时构造出最长深度后把剩下的点都插道2号点上，其余点继续插到2号点上。

#include <cstdio>
#include <iostream>
using namespace std;
int n,d,h;
int main()
{
	cin>>n>>d>>h;
	if(2*h < d)
	{
		cout<<-1<<endl;
		return 0;
	 } 
	if(h == d && h != n-1)
	{
		if(h == 1)
		{
			cout<<-1<<endl;
			return 0;
		}
		for(int i=1;i <= h;i++) cout<<i<<" "<<i+1<<endl;
	    cout<<2<<" "<<h+2<<endl;
	    for(int i=1;i <= n-h-2;i++) cout<<2<<" "<<i+h+2<<endl;
	    return 0;
	}
	for(int i=1;i <= h;i++) cout<<i<<" "<<i+1<<endl;
	int now=1;
	for(int i=1;i <= d-h;i++)
	{
		cout<<now<<" "<<i+h+1<<endl;
		now=i+h+1;
	}
	for(int i=1;i <= n-d-1;i++) cout<<1<<" "<<d+1+i<<endl;
}