#include <bits/stdc++.h>

#define FOR(a, b, c) for (int a = (b), _for = (c); a < _for; ++a)
#define REP(a, n) FOR(a, 0, n)
#define x first
#define y second
#define pb push_back
#define pii pair<int, int>
#define ll long long
#define lf double

using namespace std;

void NO(){
	puts ("-1");
	exit(0);
}

const int MAX = (int)1e6+5;

int n,m;
set<int>v[MAX];
vector<pii>out;

int bio[MAX];
int tajm;

bool dfs2(int cx,int par,int is){
	bio[cx]=++tajm;
	int ra=cx;
	int rb=par;
	if (is) swap(ra,rb);
	for (auto nx:v[cx]){
		if (nx==par) continue;
		if (bio[nx]){
			if (bio[nx]>bio[cx]){
				out.pb({cx,nx});
				is^=1;
			}
			continue;
		}
		if (dfs2(nx,cx,!is)){
			is=1;
		}
		else{
			is=0;
		}
	}
	if (par>=0){
		if (is) out.pb({ra,rb});
		else out.pb({rb,ra});
	}
	return is;
}

int main(){
	scanf ("%d%d",&n,&m);
	if (m&1) NO();
	REP(i,m){
		int a,b;
		scanf ("%d%d",&a,&b);
		a--;b--;
		v[a].insert(b);
		v[b].insert(a);
	}
	dfs2(0,-1,0);
	for (auto x:out){
		printf ("%d %d\n",x.x+1,x.y+1);
	}
	
	
	
	return 0;
}

### Prompt

Your challenge is to write 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>

#define FOR(a, b, c) for (int a = (b), _for = (c); a < _for; ++a)
#define REP(a, n) FOR(a, 0, n)
#define x first
#define y second
#define pb push_back
#define pii pair<int, int>
#define ll long long
#define lf double

using namespace std;

void NO(){
	puts ("-1");
	exit(0);
}

const int MAX = (int)1e6+5;

int n,m;
set<int>v[MAX];
vector<pii>out;

int bio[MAX];
int tajm;

bool dfs2(int cx,int par,int is){
	bio[cx]=++tajm;
	int ra=cx;
	int rb=par;
	if (is) swap(ra,rb);
	for (auto nx:v[cx]){
		if (nx==par) continue;
		if (bio[nx]){
			if (bio[nx]>bio[cx]){
				out.pb({cx,nx});
				is^=1;
			}
			continue;
		}
		if (dfs2(nx,cx,!is)){
			is=1;
		}
		else{
			is=0;
		}
	}
	if (par>=0){
		if (is) out.pb({ra,rb});
		else out.pb({rb,ra});
	}
	return is;
}

int main(){
	scanf ("%d%d",&n,&m);
	if (m&1) NO();
	REP(i,m){
		int a,b;
		scanf ("%d%d",&a,&b);
		a--;b--;
		v[a].insert(b);
		v[b].insert(a);
	}
	dfs2(0,-1,0);
	for (auto x:out){
		printf ("%d %d\n",x.x+1,x.y+1);
	}
	
	
	
	return 0;
}

```