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 repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef long long ll;
const long long INF = 1LL << 60;
typedef pair<int, int> P;

int check(int N) {
    bool flag = false;
    int k = 2, index = -1;
    repd(i, 1, N + 1) {
        if (i == N) {
            flag = true;
            index = k - 1;
            break;
        }
        i += (k - 1);
        k++;
    }

    return index;
}

int main()
{
    int N;
    cin >> N;

    int index = check(N);
    if (index != -1) {
        vector<vector<int>> ans(index + 1, vector<int>(index));
        int s = 1;
        rep(i, index) {
            rep(j, i + 1) {
                ans[j][i] = s;
                s++;
            }
        }

        s = 1;
        rep(i, index) {
            rep(j, i + 1) {
                ans[i + 1][j] = s;
                s++;
            }
        }

        cout << "Yes" << endl;
        rep(i, index + 1) {
            cout << index << " ";
            rep(j, index - 1) {
                cout << ans[i][j] << " "; 
            }
            cout << ans[i][index - 1] << endl;
        }
    }
    else {
        cout << "NO" << endl;
    }
    return 0;
}