#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m,fa[N];
vector<pair<int,int>> E[N];
pair<int,int> ans[N];
bool deg[N];

void dfs(int u,int fa) {
	for(auto x:E[u]) {
		int v=x.first,id=x.second;
		if(v==fa) continue;
		dfs(v,u);
		if(deg[v]) deg[v]=0,ans[id]=make_pair(v,u);
		else deg[u]^=1,ans[id]=make_pair(u,v);
	}
}

int getf(int x) {
	return x==fa[x]?x:fa[x]=getf(fa[x]);
}

int main() {
	cin>>n>>m;
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1,u,v;i<=m;i++) {
		u=gi(),v=gi();
		if(getf(u)!=getf(v)) fa[getf(u)]=getf(v),E[u].emplace_back(v,i),E[v].emplace_back(u,i);
		else ans[i]=make_pair(u,v),deg[u]^=1;	
	}
	dfs(1,0);
	if(deg[1]) puts("-1");
	else for(int i=1;i<=m;i++) cout<<ans[i].first<<' '<<ans[i].second<<endl;
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int N=1e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m,fa[N];
vector<pair<int,int>> E[N];
pair<int,int> ans[N];
bool deg[N];

void dfs(int u,int fa) {
	for(auto x:E[u]) {
		int v=x.first,id=x.second;
		if(v==fa) continue;
		dfs(v,u);
		if(deg[v]) deg[v]=0,ans[id]=make_pair(v,u);
		else deg[u]^=1,ans[id]=make_pair(u,v);
	}
}

int getf(int x) {
	return x==fa[x]?x:fa[x]=getf(fa[x]);
}

int main() {
	cin>>n>>m;
	for(int i=1;i<=n;i++) fa[i]=i;
	for(int i=1,u,v;i<=m;i++) {
		u=gi(),v=gi();
		if(getf(u)!=getf(v)) fa[getf(u)]=getf(v),E[u].emplace_back(v,i),E[v].emplace_back(u,i);
		else ans[i]=make_pair(u,v),deg[u]^=1;	
	}
	dfs(1,0);
	if(deg[1]) puts("-1");
	else for(int i=1;i<=m;i++) cout<<ans[i].first<<' '<<ans[i].second<<endl;
	return 0;
}

```