Lunar New Year and a Wander

D. Lunar New Year and a Wander

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Lunar New Year is approaching, and Bob decides to take a wander in a nearby park.

The park can be represented as a connected graph with n nodes and m bidirectional edges. Initially Bob is at the node 1 and he records 1 on his notebook. He can wander from one node to another through those bidirectional edges. Whenever he visits a node not recorded on his notebook, he records it. After he visits all nodes at least once, he stops wandering, thus finally a permutation of nodes a1,a2,…,an is recorded.

Wandering is a boring thing, but solving problems is fascinating. Bob wants to know the lexicographically smallest sequence of nodes he can record while wandering. Bob thinks this problem is trivial, and he wants you to solve it.

A sequence x is lexicographically smaller than a sequence y if and only if one of the following holds:

• x is a prefix of y, but x≠y (this is impossible in this problem as all considered sequences have the same length);
• in the first position where x and y differ, the sequence x has a smaller element than the corresponding element in y.

Input

The first line contains two positive integers n and m (1≤n,m≤105), denoting the number of nodes and edges, respectively.

The following m lines describe the bidirectional edges in the graph. The i-th of these lines contains two integers ui and vi (1≤ui,vi≤n), representing the nodes the i-th edge connects.

Note that the graph can have multiple edges connecting the same two nodes and self-loops. It is guaranteed that the graph is connected.

Output

Output a line containing the lexicographically smallest sequence a1,a2,…,an Bob can record.

Examples

input

Copy

3 2
1 2
1 3

output

Copy

1 2 3 

input

Copy

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

output

Copy

1 4 3 2 5 

input

Copy

10 10
1 4
6 8
2 5
3 7
9 4
5 6
3 4
8 10
8 9
1 10

output

Copy

1 4 3 7 9 8 6 5 2 10 

Note

In the first sample, Bob's optimal wandering path could be 1→2→1→3. Therefore, Bob will obtain the sequence {1,2,3}, which is the lexicographically smallest one.

In the second sample, Bob's optimal wandering path could be 1→4→3→2→3→4→1→5. Therefore, Bob will obtain the sequence {1,4,3,2,5}, which is the lexicographically smallest one.

#include<bits/stdc++.h>
#define REP(i, a, b) for(int i = (a); i <= (b); ++ i)
#define REP(j, a, b) for(int j = (a); j <= (b); ++ j)
#define PER(i, a, b) for(int i = (a); i >= (b); -- i)
using namespace std;
const int maxn=2e5+5;
template <class T>
inline void rd(T &ret){
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9'){
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}
priority_queue<int,vector<int>,greater<int> >q;
vector<int>ans;
bool vis[maxn];
int head[maxn],n,m,tot;
struct node{
    int to,nx;
}p[maxn];
void addedge(int s,int t){
     p[++tot].to=t,p[tot].nx=head[s],head[s]=tot;
}
int main()
{
    rd(n),rd(m);
    REP(i,1,m){
        int u,v;
        rd(u),rd(v);
        addedge(u,v),addedge(v,u);
    }
    q.push(1);
    while(!q.empty()){
         int cur=q.top();
         q.pop();
         if(vis[cur])continue;
         vis[cur]=true;
         ans.push_back(cur);
         for(int i=head[cur];i;i=p[i].nx){
              int to=p[i].to;
              if(vis[to])continue;
             // vis[to]=1;
              q.push(to);
         }
    }
    for(int i=0;i<ans.size();i++)cout<<ans[i]<<' ';
    return 0;
}