Problem:
Score : 500 points

Problem Statement

You are given an integer N. Determine if there exists a tuple of subsets of \{1,2,...N\}, (S_1,S_2,...,S_k), that satisfies the following conditions:

Each of the integers 1,2,...,N is contained in exactly two of the sets S_1,S_2,...,S_k.

Any two of the sets S_1,S_2,...,S_k have exactly one element in common.

If such a tuple exists, construct one such tuple.

Constraints

1 \leq N \leq 10^5

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If a tuple of subsets of \{1,2,...N\} that satisfies the conditions does not exist, print No.
If such a tuple exists, print Yes first, then print such subsets in the following format:

k
|S_1| S_{1,1} S_{1,2} ... S_{1,|S_1|}
:
|S_k| S_{k,1} S_{k,2} ... S_{k,|S_k|}

where S_i={S_{i,1},S_{i,2},...,S_{i,|S_i|}}.

If there are multiple such tuples, any of them will be accepted.

Sample Input 1

3

Sample Output 1

Yes
3
2 1 2
2 3 1
2 2 3

It can be seen that (S_1,S_2,S_3)=(\{1,2\},\{3,1\},\{2,3\}) satisfies the conditions.

Sample Input 2

4

Sample Output 2

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>

#define INF 1000000021
#define MOD 1000000007
#define pb push_back
#define sqr(a) (a)*(a)
#define M(a, b) make_pair(a,b)
#define T(a, b, c) make_pair(a, make_pair(b, c))
#define F first
#define S second
#define all(x) (x.begin(), x.end())
#define deb(x) cerr << #x << " = " << x << '\n'
#define N 222222

using namespace std;
//using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;

const ld pi = 2 * acos(0.0);
template<class T> bool umin(T& a, T b){if(a>b){a=b;return 1;}return 0;}
template<class T> bool umax(T& a, T b){if(a<b){a=b;return 1;}return 0;}
template<class T, class TT> bool pal(T a, TT n){int k=0;for(int i=0;i<=n/2;i++){if(a[i]!=a[n-i-1]){k=1;break;}}return k?0:1;}

//int month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

int main()
{
	int n;
	cin >> n;
	if(n == 1)
		cout << "Yes\n2\n1 1\n1 1";
	else
	if(n == 2)
		cout << "Yes\n2\n2 1 2\n2 2 1";
	else
	if(n == 3)
		cout << "Yes\n3\n2 1 2\n2 3 1\n2 2 3";
	else
		cout << "No";





	





	getchar();
	getchar();
	return 0;
	//ios::sync_with_stdio(false);
	//cin.tie(0);
}