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 <cstdio>
#include <cstdlib>

#include <vector>
#include <iostream>
#include <algorithm>
typedef long long LL;
typedef unsigned long long ULL;
const int MAXN = 100005;
const int P = 1000000007;
const int INF = 1000000007;
const double eps = 1e-6;
using namespace std;
inline int getint() {
    int r = 0; bool b = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') b = false; c = getchar(); }
    while (c >= '0' && c <= '9') { r = (r<<1)+(r<<3) + c - '0'; c = getchar(); }
    return b ? r : -r;
}
vector<int> v[450];
int n, k = -1;
int main() {
    n = getint();
    for (int i = 1; i * (i - 1) <= n * 2; ++i) {
        if (i * (i - 1) == n * 2) {
            k = i;
            break;
        }
    }
    if (k == -1) puts("No");
    else {
        int tot = 0;
        for (int i = 1; i < k; ++i)
            for (int j = i + 1; j <= k; ++j) {
                ++tot;
                v[i].push_back(tot);
                v[j].push_back(tot);
            }
        printf("Yes\n%d\n", k);
        for (int i = 1; i <= k; ++i) {
            printf("%d", k - 1);
            for (auto &j : v[i])
                printf(" %d", j);
            putchar('\n');
        }
    }
}