Problem: You are given a simple connected undirected graph with N vertices and M edges. The vertices are numbered 1 to N, and the i-th edge connects Vertex A_i and Vertex B_i. Takahashi will assign one of the two possible directions to each of the edges in the graph to make a directed graph. Determine if it is possible to make a directed graph with an even number of edges going out from every vertex. If the answer is yes, construct one such graph.

Constraints

* 2 \leq N \leq 10^5
* N-1 \leq M \leq 10^5
* 1 \leq A_i,B_i \leq N (1\leq i\leq M)
* The given graph is simple and connected.

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
:
A_M B_M


Output

If it is impossible to assign directions to satisfy the requirement, print -1. Otherwise, print an assignment of directions that satisfies the requirement, in the following format:


C_1 D_1
:
C_M D_M


Here each pair (C_i, D_i) means that there is an edge directed from Vertex C_i to Vertex D_i. The edges may be printed in any order.

Examples

Input

4 4
1 2
2 3
3 4
4 1


Output

1 2
1 4
3 2
3 4


Input

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


Output

-1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

const int N=1000005;
int f[N],pa[N],n,m,a,b;
vector<pii> ans;
map<pii,int> E;
vi e[N];

inline int getpa(int x){
	return pa[x]==x?x:pa[x]=getpa(pa[x]);
}

void dfs(int u,int fa){
	for(auto v:e[u])
		if(v!=fa){
			dfs(v,u);
			if(f[v]){
				f[u]^=1;
				f[v]^=1;
				if(E[mp(u,v)])ans.pb(mp(v,u));
				else ans.pb(mp(u,v));
			}
			else{
				if(E[mp(u,v)])ans.pb(mp(u,v));
				else ans.pb(mp(v,u));
			}
		}
}

int main(){
	cin>>n>>m;
	rep(i,1,n)pa[i]=i;
	rep(i,1,m){
		cin>>a>>b;
		E[mp(a,b)]=1;
		if(getpa(a)==getpa(b)){
			f[a]^=1;
			ans.pb(mp(a,b));
		}
		else{
			f[a]^=1;
			e[a].pb(b);
			e[b].pb(a);
			pa[getpa(a)]=getpa(b);
		}
	}
	dfs(1,0);
	if(f[1])return puts("-1"),0;
	for(auto x:ans)
		printf("%d %d\n",x.fi,x.se);
	return 0;
}