#ifndef ONLINE_JUDGE
#define _GLIBCXX_DEBUG
#endif
#include<bits/stdc++.h>

#define PI 3.14159265359
using namespace std;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define bit(n,k) (((ll)n>>(ll)k)&1) /*nのk bit目*/
#define pb push_back
#define eb emplace_back
#define SZ(x) ((ll)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
const long long INF= 1e+18+1;
//long long INF=(1LL<<31)-1;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll> >vvl;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> T;
const ll MOD=1000000007LL;
//ll MOD=1000000007LL;
//const ll MOD=998244353LL;
//const ll MOD=1777777777LL;
//const ll MAX_V=114514LL;
//const ll MAX = 500010LL;
const ll mod=MOD;
string abc="abcdefghijklmnopqrstuvwxyz";
string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
vl dx={0,0,1,-1};
vl dy={1,-1,0,0};

//素因数分解O(√n)
map<ll,ll>prime_factor(ll n){
  map<ll,ll>res;
  for(ll i=2;i*i<=n;i++){
    while(n%i==0){
      res[i]++;
      n/=i;
    }
  }
  if(n!=1)res[n]=1;
  return res;
}

const ll MAX = 2000010;
long long fac[MAX], finv[MAX], inv[MAX];
//finvが階乗の逆元

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll modpow(ll a, ll n, ll mod) {
  
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
/*ll N=2000010;
vl arr(N);
void Eratosthenes(){
	for(ll i = 0; i < N; i++){
		arr[i] = 1;
	}
        arr[1]=0;
	for(ll i = 2; i < sqrt(N); i++){
		if(arr[i]){
			for(ll j = 0; i * (j + 2) < N; j++){
				arr[i *(j + 2)] = 0;
			}
		}
	}
}*/
//素数判定O(√n)
bool is_prime(ll n){
  for(ll i=2;i*i<=n;i++){
    if(n%i==0)return false;
  }
  return n!=1;
}

/* SegTree<X>(n,fx,ex): モノイド(集合X, 二項演算fx, 単位元ex)についてサイズnで構築
    set(int i, X x), build(): i番目の要素をxにセット。まとめてセグ木を構築する。O(n)
    update(i,x): i 番目の要素を x に更新。O(log(n))
    query(a,b):  [a,b) 全てにfxを作用させた値を取得。O(log(n))
*/
template <typename X>
struct SegTree {
    using FX = function<X(X, X)>;
    int n;
    FX fx;
    const X ex;
    vector<X> dat;
    SegTree(int n_, FX fx_, X ex_) : n(), fx(fx_), ex(ex_), dat(n_ * 4, ex_) {
        int x = 1;
        while (n_ > x) {
            x *= 2;
        }
        n = x;
    }

    void set(int i, X x) { dat[i + n - 1] = x; }
    void build() {
        for (int k = n - 2; k >= 0; k--) dat[k] = fx(dat[2 * k + 1], dat[2 * k + 2]);
    }

    void update(int i, X x) {
        i += n - 1;
        dat[i] = x;
        while (i > 0) {
            i = (i - 1) / 2;  // parent
            dat[i] = fx(dat[i * 2 + 1], dat[i * 2 + 2]);
        }
    }

    // the minimum element of [a,b)
    X query(int a, int b) { return query_sub(a, b, 0, 0, n); }
    X query_sub(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) {
            return ex;
        } else if (a <= l && r <= b) {
            return dat[k];
        } else {
            X vl = query_sub(a, b, k * 2 + 1, l, (l + r) / 2);
            X vr = query_sub(a, b, k * 2 + 2, (l + r) / 2, r);
            return fx(vl, vr);
        }
    }
    /* debug */
    inline X operator[](int a) { return query(a, a + 1); }
    void print() {
        for (int i = 0; i < 2 * n - 1; ++i) {
            cout << (*this)[i];
            if (i != n) cout << ",";
        }
        cout << endl;
    }
    /*
    使用例
    auto fx=[](int x1,int x2)->int{return max(x1,x2);};
    ll ex=0;
    SegTree<ll>rmq(n,fx,ex);*/
};

int main(){
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  cout << fixed << setprecision(10);
  /*--------------------------------*/
  
  ll n;cin>>n;
  ll k=-1;
  for(ll i=1;i<=1000;i++){
      if(i*(i-1)/2==n)k=i;
  }
  if(k==-1){cout<<"No"<<endl;return 0;}
  cout<<"Yes"<<endl;
  cout<<k<<endl;
  vvl ans(k);
  queue<ll>q;
  for(ll i=1;i<=n;i++)q.push(i);
  for(ll i=0;i<k-1;i++){
      for(ll j=i+1;j<k;j++){
          ans[i].pb(q.front());
          ans[j].pb(q.front());
          q.pop();
      }
  }
  rep(i,k){
      cout<<2*n/k<<" ";
      rep(j,2*n/k)cout<<ans[i][j]<<" \n"[j+1==2*n/k];
  }
}



Measured memory_bytes: 5208