CodeForces-329C（div1）：Graph Reconstruction（随机&构造）

I have an undirected graph consisting of n nodes, numbered 1 through n. Each node has at most two incident edges. For each pair of nodes, there is at most an edge connecting them. No edge connects a node to itself.

I would like to create a new graph in such a way that:

• The new graph consists of the same number of nodes and edges as the old graph.
• The properties in the first paragraph still hold.
• For each two nodes u and v, if there is an edge connecting them in the old graph, there is no edge connecting them in the new graph.

Help me construct the new graph, or tell me if it is impossible.

Input

The first line consists of two space-separated integers: n and m (1 ≤ m ≤ n ≤ 10⁵), denoting the number of nodes and edges, respectively. Then m lines follow. Each of the m lines consists of two space-separated integers u and v (1 ≤ u, v ≤ n; u ≠ v), denoting an edge between nodes u and v.

Output

If it is not possible to construct a new graph with the mentioned properties, output a single line consisting of -1. Otherwise, output exactly m lines. Each line should contain a description of edge in the same way as used in the input format.

Examples

Input

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

Output

1 4
4 6
1 6
2 7
7 5
8 5
2 8

Input

3 2
1 2
2 3

Output

-1

Input

5 4
1 2
2 3
3 4
4 1

Output

1 3
3 5
5 2
2 4

题意：给定N点M边的无向简单图，满足每个点最多度数为2。现在叫你重新构图，使得新图满足原来的性质，而且新图的边和原图的边没有相同的。

思路：因为度数最多为2，我们去随机构造一个链（最坏的情况下是个环），如果新图和就图没有交集，则ok。

#include<bits/stdc++.h>
#define pii pair<int,int>
#define mp make_pair
#define f first
#define s second
using namespace std;
const int maxn=;
map<pii,int>ex;
pii p[maxn];
int N,M,ans[maxn];
bool solve()
{
    random_shuffle(ans+,ans+N+); ans[]=ans[N]; int cnt=;
    for(int i=;i<=N;i++)
      if(!ex[mp(ans[i-],ans[i])]) {
        cnt++; p[cnt]=mp(ans[i-],ans[i]);
    }
    if(cnt<M) return false;
    return true;
}
int main()
{
    int T,u,v,i,j;
    scanf("%d%d",&N,&M);
    for(i=;i<=M;i++){
        scanf("%d%d",&u,&v);
        ex[mp(u,v)]=; ex[mp(v,u)]=;
    }
    for(i=;i<=N;i++) ans[i]=i;
    T=/M;
    while(T--){
        if(solve()) {
           for(i=;i<=M;i++) printf("%d %d\n",p[i].f,p[i].s);
           return ;
        }
    }
    puts("-1");
    return ;
}