1003E. Tree Constructing(构造)

You are given three integers $n$, $d$ and $k$

.

Your task is to construct an undirected tree on $n$

vertices with diameter $d$ and degree of each vertex at most $k$

, or say that it is impossible.

An undirected tree is a connected undirected graph with $n-1$

edges.

Diameter of a tree is the maximum length of a simple path (a path in which each vertex appears at most once) between all pairs of vertices of this tree.

Degree of a vertex is the number of edges incident to this vertex (i.e. for a vertex $u$

it is the number of edges $(u,v)$ that belong to the tree, where $v$

is any other vertex of a tree).

Input

The first line of the input contains three integers $n$

, $d$ and $k$ ($1\le n,d,k\le 4\cdot {10}^5$

).

Output

If there is no tree satisfying the conditions above, print only one word "NO" (without quotes).

Otherwise in the first line print "YES" (without quotes), and then print $n-1$

lines describing edges of a tree satisfying the conditions above. Vertices of the tree must be numbered from $1$ to $n$

. You can print edges and vertices connected by an edge in any order. If there are multiple answers, print any of them.1

Examples

Input

Copy

6 3 3

Output

Copy

YES
3 1
4 1
1 2
5 2
2 6

Input

Copy

6 2 3

Output

Copy

NO

Input

Copy

10 4 3

Output

Copy

YES
2 9
2 10
10 3
3 1
6 10
8 2
4 3
5 6
6 7

Input

Copy

8 5 3

Output

Copy

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

题意

让你构造一个节点数为n，直径为d，且所有点的度数均小于k的树。

题解

我们先用1~(n+1)把直径做出来，然后用dfs去添加其他结点，我们可以在第i个结点去挂上一棵深度不超过min(i-1,d-i+1)的树，注意在生成这棵树的时候每个结点的度数不能超过k。总结点数==n时，及时结束。

那么什么时候无解呢，①节点数小于等于直径（这样直径都搞不出来），②直径＞1且每个点的度数不能超过2，③n个结点没有全部挂上去。

#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
using namespace std;
int n,d,k;
int id;
vector< pair<int,int> >ans;
void DFS(int u,int dep,int maxd){
	if(dep==maxd) return;
	for(int i=0;i<k-1-(dep==0) && id<n;i++){
		ans.push_back(make_pair(u,++id));
		DFS(id,dep+1,maxd);
	}
}
int main(){
	cin>>n>>d>>k;
	if(n<=d||d>1&&k<2){
		cout<<"NO"<<endl;
		return 0;
	}
	for(int i=1;i<=d;i++)
	   ans.push_back(make_pair(i,i+1));
	id=d+1;
	for(int i=2;i<=d;i++) DFS(i,0,min(i-1,d-i+1));
	if(id<n) cout<<"NO"<<endl;
	else{
		cout<<"YES"<<endl;
		for(int i=0;i<ans.size();i++)
		   cout<<ans[i].first<<" "<<ans[i].second<<endl;
	}
	return 0;
}