#include <bits/stdc++.h>
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
//#define mod 998244353
//#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;

template <class T>
void chkmax(T &x,T y){
	if(x < y) x = y;
}

template <class T>
void chkmin(T &x,T y){
	if(x > y) x = y;
}

inline int popcnt(int x){
	return __builtin_popcount(x);
}

inline int ctz(int x){
	return __builtin_ctz(x);
}


/*ll power(ll p,int k = mod - 2){
	ll ans = 1;
	while(k){
		if(k % 2 == 1) ans = ans * p % mod;
		p = p * p % mod;
		k /= 2;	
	}
	return ans;
}*/

int cnt;
int head[350005];
struct eg{
	int to,nxt,w;
}edge[2200005];

void make(int u,int v,int w){
	edge[++cnt].to = v;
	edge[cnt].w = w;
	edge[cnt].nxt = head[u];
	head[u] = cnt;
}
void mk(int u,int v,int w){
//	printf("mk %d->%d %d\n",u,v,w);
	make(u,v,w);make(v,u,0);
}
int s,t;

int dis[350005],cur[350005];
queue <int> Q;
int bfs(){
	fill(dis,dis + t + 1,inf);
	dis[s] = 0;
	Q.push(s);
	while(!Q.empty()){
		int u = Q.front();
		Q.pop();
		for(int i = head[u];i;i = edge[i].nxt){
			int v = edge[i].to;
			if(!edge[i].w) continue;
			if(dis[v] > dis[u] + 1){
				dis[v] = dis[u] + 1;
				Q.push(v);
			} 
		}
	}
	return dis[t] != inf;
}

ll dfs(int u,ll flow){
	if(u == t || !flow) return flow;
	ll sum = 0,temp;
	for(int i = cur[u];i;i = cur[u]){
		cur[u] = edge[i].nxt;
		int v = edge[i].to;
		if(!edge[i].w || dis[v] != dis[u] + 1) continue;
		temp = dfs(v,min(flow,1ll * edge[i].w));
		sum += temp;
		flow -= temp;
		edge[i].w -= temp;
		edge[i ^ 1].w += temp;
		if(!flow) return sum;
	}
	return sum;
}

ll dinic(){
	ll ans = 0;
	while(bfs()){
		copy(head,head + t + 1,cur);
		ans += dfs(s,linf);
//		printf("%lld\n",ans);
	}
	return ans;
}

ll n;
int ans[505];
void resolve(int k){
	cnt = 1;
	s = 0;t = 2 * k + 1;
	fill(head,head + t + 1,0);
	rep(u,1,k) mk(s,u,1);
	rep(v,1,k) mk(k + v,t,1);
	rep(u,1,k){
		rep(v,1,k) if((n + 1 - v) % u == 0) mk(u,k + v,1);
	}
	ans[k] = (dinic() == k);
}

int idx[65];
const int lim = 300;
vector <int> answer;
int main(){
	scanf("%lld",&n);
	if(n <= 2 * lim){
		rep(k,0,n / 2){
			resolve(k);
			ans[n - k] = ans[k];
		}
		printf("2\n%d ",n + 1);
		rep(k,0,n) printf("%d ",ans[k]);
		printf("\n");
		return 0;
	}
	rep(k,0,lim) resolve(k);
	printf("62\n");
	idx[0] = 0;
	rep(k,1,60){
		printf("2 %d %d\n",idx[k - 1],idx[k - 1]);
		idx[k] = k + 1;
	}
	rep(k,0,lim) answer.eb(ans[k]);
	ll sum = n - 2 * lim - 1;
	per(k,60,0){
		if(sum >= (1ll << k)){
			answer.eb(idx[k]);
			sum -= 1ll << k;
		}
	}
	per(k,lim,0) answer.eb(ans[k]);
	printf("%d\n",SZ(answer));
	for(int p:answer) printf("%d ",p);
	printf("\n");
	//[0,lim] 和 [n-lim,n] 范围内的都好了,再构造 n-2lim-1 个 0
	return 0;
}