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>

using namespace std;

#define rep(i, x)   for (__typeof(x) i = 0; i < x; i++)
#define mp             make_pair
#define pb             push_back
#define fi             first
#define se             second
#define sz(x)          (int)((x).size())
#define all(x)         (x).begin(), (x).end()
#define sci(x)         int x;       scanf("%d", &x);
#define scii(x, y)     int x, y;    scanf("%d %d", &x, &y);
#define sciii(x, y, z) int x, y, z; scanf("%d %d %d", &x, &y, &z);
#define TC(x)          sci(x); while(x --)
#define eprintf(...)   fprintf(stderr, __VA_ARGS__)
#define debug(x)       { cerr << #x << " = " << x << endl; }
#define repi(i, x)  for (__typeof(x) i = x-1; i >= 0; i --)
#define fore(itr, x)   for (__typeof(x.begin()) itr = x.begin(); itr != x.end(); itr ++)
#define forei(itr, x)  for (__typeof(x.end()) itr = x.end() - 1; itr != x.begin() - 1; itr --)
typedef long long      ll;
typedef pair<int, int> ii;
typedef pair<ii, int>  iii;
typedef vector<int>    vi;
typedef vector<ii>     vii;
typedef vector<iii>    viii;
typedef vector<vector<int> > vvi;
const   int            infi = numeric_limits<int>::max();
const   double         eps = 0;
const   int            ms  = 0;

ll solve(){
    ll N;
    cin >> N;
    if(N<=2)return -1;
    ll n = 3;
    for(; n*(n-1)/2 <= N; n++){
    }
    n--;
    if(n*(n-1)/2 == N)return n;
    return -1;
}

int main()
{
    ios::sync_with_stdio(false);//endl->"\n"
    cin.tie(0);
    ll n = solve();
    if(n>0){
        cout << "Yes" << "\n";
        vector<vector<ll>> s(n, vector<ll>(0));
        ll k = 1;
        for(ll i = 0; i < n; i++){
            for(ll j = i+1; j < n; j++){
                s[i].pb(k);
                s[j].pb(k);
                k++;
            }

        }
        rep(i, n){
            cout << n-1 << " ";
            rep(j, n-1-1){
                cout << s[i][j] << " ";
            } 
            cout << s[i][n-1-1] << "\n";
        }
    } else cout << "No" << "\n";
    return 0;
}