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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define INF 999999999
#define m0(x) memset(x,0,sizeof(x))
#define fill(x,y) memset(x,y,sizeof(x))

using namespace std;


int dy[4] = { 0,0,1,-1 };
int dx[4] = { 1,-1,0,0 };
int dxx[8] = { 0,0,1,1,1,-1,-1,-1 };
int dyy[8] = { 1,-1,0,1,-1,0,1,-1 };




ll gcd(ll x, ll y) {
	ll m = max(x, y), n = min(x, y);
	if (m%n == 0)return n;
	else return gcd(m%n, n);
}
ll lcm(ll x, ll y) {
	return x / gcd(x, y)*y;
}

ll myPow(ll x, ll n, ll m) {
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return myPow(x * x % m, n / 2, m);
	else
		return x * myPow(x, n - 1, m) % m;
}

long long nCr(int n, int r) {
	if (r > n / 2) r = n - r; // because C(n, r) == C(n, n - r)
	long long ans = 1;
	int i;

	for (i = 1; i <= r; i++) {
		ans *= n - r + i;
		ans /= i;
	}

	return ans;
}


const int MOD = 1000000007;


ll A[500];
ll ans[400][400];
int main() {
	REP(i, 400) {
		A[i] = i * (i + 1);
	}
	int N;
	cin >> N;
	bool found = false;
	REP(i, 400) {
		if (A[i] == N * 2) {
			found = true;
			break;
		}
	}
	if (!found) {
		cout << "No" << endl;
		return 0;
	}
	ll m=0, n=0;
	REP(i, 400) {
		if (i*(i + 1) == N * 2) {
			m = i;
			n = i + 1;
			break;
		}
	}
	ll cnt = 1;
	REP(i, m) {
		FOR(j, i, m) {
			ans[i][j] = cnt;
			cnt++;
		}
	}
	ll cnt2 = 1;
	REP(i, m) {
		FOR(j, i+1, m + 1) {
			ans[j][i] = cnt2;
			cnt2++;
		}
	}
	cout << "Yes" << endl;
	REP(i, n) {
		REP(j, m) {
			cout << ans[i][j] << " ";
		}
		cout << endl;
	}
}