#include<bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define print(A,x,n) rep(I,0,n){cout<<(I ? " ":"")<<A[I]x;}cout<<endl;
#define pprint(A,y,m,n) rep(J,0,m){print(A[J],y,n);}
const long mod=998244353;
const int size=1e5;
const int inf=1e9;
int main(){
	int N,M; cin>>N>>M;
	if(M % 2 == 1){ cout<<-1<<endl; return 0; }
	int a,b;
	set<int> g[size],h[size];
	int deg[size] = {};
	rep(i,0,M){
		cin>>a>>b; a--; b--;
		g[a].insert(b);
		g[b].insert(a);
		deg[a]++; deg[b]++;
	}
	set<pair<int,int>> st;
	rep(i,0,N) st.insert({deg[i], i});
	rep(i,0,N){
		pair<int,int> p = *st.begin();
		st.erase(st.begin());
		int v = p.second;
		bool flag = false;
		if(h[v].size() % 2 == 1) flag = true;
		for(int u:g[v]){
			if(h[v].count(u) || h[u].count(v)) continue;
			if(flag){ h[v].insert(u); flag = false; }
			else h[u].insert(v);
			st.erase(st.find({deg[u], u}));
            deg[v]--; deg[u]--;
            st.insert({deg[u], u});
		}
	}
	rep(i,0,N){
		for(int j:h[i]) cout<<i + 1<<" "<<j + 1<<endl;
	}
}
### Prompt

Your task is to create a cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define print(A,x,n) rep(I,0,n){cout<<(I ? " ":"")<<A[I]x;}cout<<endl;
#define pprint(A,y,m,n) rep(J,0,m){print(A[J],y,n);}
const long mod=998244353;
const int size=1e5;
const int inf=1e9;
int main(){
	int N,M; cin>>N>>M;
	if(M % 2 == 1){ cout<<-1<<endl; return 0; }
	int a,b;
	set<int> g[size],h[size];
	int deg[size] = {};
	rep(i,0,M){
		cin>>a>>b; a--; b--;
		g[a].insert(b);
		g[b].insert(a);
		deg[a]++; deg[b]++;
	}
	set<pair<int,int>> st;
	rep(i,0,N) st.insert({deg[i], i});
	rep(i,0,N){
		pair<int,int> p = *st.begin();
		st.erase(st.begin());
		int v = p.second;
		bool flag = false;
		if(h[v].size() % 2 == 1) flag = true;
		for(int u:g[v]){
			if(h[v].count(u) || h[u].count(v)) continue;
			if(flag){ h[v].insert(u); flag = false; }
			else h[u].insert(v);
			st.erase(st.find({deg[u], u}));
            deg[v]--; deg[u]--;
            st.insert({deg[u], u});
		}
	}
	rep(i,0,N){
		for(int j:h[i]) cout<<i + 1<<" "<<j + 1<<endl;
	}
}
```