#include<bits/stdc++.h>
using namespace std;
int gi() {
	char ch='!'; int res=0, f=0;
	while(ch<'0' || ch>'9') { ch=getchar(); if(ch == '-') f=1; }
	while(ch>='0' && ch<='9') res=res*10+ch-'0', ch=getchar();
	return f?-res:res;
}
const int N=1e5+100;
int n,m,fa[N],deg[N],head[N],ce;
struct edge { int to,nxt; } e[N<<1];
void lnk(int a,int b) { e[++ce]=(edge){b,head[a]}, head[a]=ce; }
int getfa(int x) { return x == fa[x]?x:fa[x]=getfa(fa[x]); }
void dfs(int u,int lst) {
	for(int i=head[u];i;i=e[i].nxt) {
		int v=e[i].to;
		if(v == lst) continue;
		dfs(v,u);
	}
	if(!lst) return;
	if(deg[u]&1)
		++deg[u], printf("%d %d\n",u,lst);
	else
		++deg[lst], printf("%d %d\n",lst,u);
}
int main() {
	n=gi(), m=gi();
	if(m&1) { puts("-1"); return 0; }
	for(int i=1;i<=n;++i) fa[i]=i;
	for(int i=1,u,v,fu,fv;i<=m;++i) {
		u=gi(), v=gi(), fu=getfa(u), fv=getfa(v);
		if(fu == fv) { printf("%d %d\n",u,v); ++deg[u]; continue; }
		fa[fu]=fv, lnk(u,v), lnk(v,u);
	}
	dfs(1,0);
	return 0;
}
### Prompt

Create a solution in cpp for 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;
int gi() {
	char ch='!'; int res=0, f=0;
	while(ch<'0' || ch>'9') { ch=getchar(); if(ch == '-') f=1; }
	while(ch>='0' && ch<='9') res=res*10+ch-'0', ch=getchar();
	return f?-res:res;
}
const int N=1e5+100;
int n,m,fa[N],deg[N],head[N],ce;
struct edge { int to,nxt; } e[N<<1];
void lnk(int a,int b) { e[++ce]=(edge){b,head[a]}, head[a]=ce; }
int getfa(int x) { return x == fa[x]?x:fa[x]=getfa(fa[x]); }
void dfs(int u,int lst) {
	for(int i=head[u];i;i=e[i].nxt) {
		int v=e[i].to;
		if(v == lst) continue;
		dfs(v,u);
	}
	if(!lst) return;
	if(deg[u]&1)
		++deg[u], printf("%d %d\n",u,lst);
	else
		++deg[lst], printf("%d %d\n",lst,u);
}
int main() {
	n=gi(), m=gi();
	if(m&1) { puts("-1"); return 0; }
	for(int i=1;i<=n;++i) fa[i]=i;
	for(int i=1,u,v,fu,fv;i<=m;++i) {
		u=gi(), v=gi(), fu=getfa(u), fv=getfa(v);
		if(fu == fv) { printf("%d %d\n",u,v); ++deg[u]; continue; }
		fa[fu]=fv, lnk(u,v), lnk(v,u);
	}
	dfs(1,0);
	return 0;
}
```