CodeForces 657A Bear and Forgotten Tree 3

CodeForces 657A Bear and Forgotten Tree 3

题目链接：

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 ofn vertices and n  -  1 edges. Vertices are numbered1 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 valuesn, d andh:

• 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 vertex1 and after that its height was h. In other words, h was the biggest distance between vertex1 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 vertex1, 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. Printn - 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.

思路：结果不唯一，构造算法。我的构造过程是，先按深度一个一个节点构造，再按直径一个一个节点构造，最后将剩下的节点放在一个节点上即可。注意几种特殊情况，一种是高和直径为1，而节点数不为2；另一种是高和直径相同。

直接附上AC代码：

#include <bits/stdc++.h>
using namespace std;
int n, d, h;int main(){#ifdef LOCAL//freopen(&#", "r", stdin);//freopen(&#", "w", stdout);#endifios::sync_with_stdio(false);cin.tie(0);cin >> n >> d >> h;if (2*h<d || d<h)       // 满足此条件，显然不可能cout << "-1" << endl;else if (n == 2){if (d == 1)cout << "1 2" << endl;elsecout << "-1" << endl;}else if (d == h){if (n != 2 && d==1 && h==1){        // 节点数太多，不科学cout << "-1" << endl;return 0;}for (int i=1; i<=d; ++i)cout << i << " " << i+1 << endl;for (int i=d+2; i<=n; ++i)          // 高度和直径一样，换个节点构造cout << "2 " << i << endl;}else{for (int i=1; i<=h; ++i)            // 构造高度cout << i << " " << i+1 << endl;int cnt = h+2;if (h < n-1)cout << "1 " << cnt << endl;for (int i=0; i<d-h-1; ++i, ++cnt)  // 构造直径cout << cnt << " " << cnt+1 << endl;cnt += 1;for (int i=0; i<n-1-d; ++i, ++cnt)  // 构造剩余节点cout << "1 " << cnt << endl;}return 0;
}