# Ehab’s Last Theorem

It’s the year 5555. You have a graph, and you want to find a long cycle and a huge independent set, just because you can. But for now, let’s just stick with finding either.

Given a connected graph with $n$ vertices, you can choose to either:

• find an independent set that has exactly $\lceil\sqrt{n}\rceil$ vertices.
• find a simple cycle of length at least $\lceil\sqrt{n}\rceil$.

An independent set is a set of vertices such that no two of them are connected by an edge. A simple cycle is a cycle that doesn’t contain any vertex twice. I have a proof you can always solve one of these problems, but it’s too long to fit this margin.

Input

The first line contains two integers $n$ and $m$ ($5 \le n \le 10^5$, $n-1 \le m \le 2 \cdot 10^5$) — the number of vertices and edges in the graph.

Each of the next $m$ lines contains two space-separated integers $u$ and $v$ ($1 \le u,v \le n$) that mean there’s an edge between vertices $u$ and $v$. It’s guaranteed that the graph is connected and doesn’t contain any self-loops or multiple edges.

Output

If you choose to solve the first problem, then on the first line print “1”, followed by a line containing $\lceil\sqrt{n}\rceil$ distinct integers not exceeding $n$, the vertices in the desired independent set.

If you, however, choose to solve the second problem, then on the first line print “2”, followed by a line containing one integer, $c$, representing the length of the found cycle, followed by a line containing $c$ distinct integers integers not exceeding $n$, the vertices in the desired cycle, in the order they appear in the cycle.

Examples

input

6 6
1 3
3 4
4 2
2 6
5 6
5 1


output

1
1 6 4

input

6 8
1 3
3 4
4 2
2 6
5 6
5 1
1 4
2 5


output

2
4
1 5 2 4

input

5 4
1 2
1 3
2 4
2 5


output

1
3 4 5 

Note

In the first sample:

Notice that you can solve either problem, so printing the cycle $2-4-3-1-5-6$ is also acceptable.

In the second sample:

Notice that if there are multiple answers you can print any, so printing the cycle $2-5-6$, for example, is acceptable.

In the third sample:

Solution:

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

const int N=101000;
int dep[N],vis[N],par[N];
int n,m,u,v,d,cnt[N];
VI e[N];
void dfs(int u) {
vis[u]=1;
for (auto v:e[u]) {
if (vis[v]&&dep[v]-dep[u]>=d-1) {
VI cyc;
int x=v;
while (x!=u) {
cyc.pb(x); x=par[x];
}
cyc.pb(u);
assert(SZ(cyc)>=d);
puts("2");
printf("%d\n",SZ(cyc));
for (auto u:cyc) printf("%d ",u); puts("");
exit(0);
} else if (!vis[v]) {
dep[v]=dep[u]+1;
par[v]=u;
dfs(v);
}
}
}
int main() {
scanf("%d%d",&n,&m);
rep(i,0,m) {
scanf("%d%d",&u,&v);
e[u].pb(v);
e[v].pb(u);
}
d=sqrt(n)-1;
while (d*d<n) d+=1;
dfs(1);
rep(i,1,n+1) {
cnt[dep[i]%(d-1)]++;
}
int p=max_element(cnt,cnt+d-1)-cnt;
VI ind;
rep(i,1,n+1) if (dep[i]%(d-1)==p) ind.pb(i);
ind.resize(d);
puts("1");
for (auto x:ind) printf("%d ",x);
puts("");
}