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: Accepted)

#include <bits/stdc++.h>
#pragma GCC optimize("03")
#define ll long long
#define ld long double
#define fi first
#define se second

using namespace std;

int sz[500100], n, sol[1000][1000], nr[4000100], vf;
vector <int> v;
bool ok[500100];

int main() {
//	ifstream cin("tst.in");
	ofstream out("tst.out");
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	int p = 1;
	int rat = 2;
	while (p <= 200000) {
		ok[p] = 1;
		sz[p] = rat;
		p += rat;
		rat++;
	}
	cin >> n;
	if (!ok[n]) {
		cout << "No";
		return 0;
	}
	int col = sz[n];
	int row = 2 * n / col;
	int pos = 0;
	for (int i = 1; i <= 400000; i++) {
		nr[++pos] = i;
		nr[++pos] = i;
	}
	int l = 2;
	int h = 1;
	while (l <= 1000) {
		int p1, p2;
		p1 = p2 = 1;
		for (int i = 1; i <= h; i++) {
			int it = nr[++vf];
			sol[h][p1++] = it;
			int it2 = nr[++vf];
			sol[p2++][l] = it2;
		}
		l++;
		h++;
	}	
	cout << "Yes\n" << col << '\n';
	for (int i = 1; i <= col; i++) {
		cout << row << ' ';
		for (int j = 1; j <= row; j++)
			cout << sol[j][i] << ' ';
		cout << '\n';
	} 
	for (int i = 1; i <= 30; i++, out << '\n')
		for (int j = 1; j <= 31; j++)
			out << sol[i][j] << ' ';
	return 0;
}