#include<bits/stdc++.h>
#define ll long long
#define clr(x,i) memset(x,i,sizeof(x))
using namespace std;
const int N=100005;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
vector<int> G[N];
int n,m,fa[N],dfn[N],f[N],tot;
void dfs(int u,int pre)
{
	fa[u]=pre; dfn[u]=++tot;
	for(auto v:G[u]){
		if(v==pre) continue;
		if(fa[v] && dfn[v]<dfn[u]) f[u]^=1,printf("%d %d\n",u,v);
		else if(!fa[v]) dfs(v, u);
	}
}
void dfs2(int u)
{
	for(auto v:G[u]){
		if(v==fa[u] || fa[v]!=u) continue;
		dfs2(v);
	}
	if(u==1) return;
	if(!f[u]) printf("%d %d\n",fa[u],u),f[fa[u]]^=1;
	else printf("%d %d\n",u,fa[u]),f[u]^=1;
}
int main()
{
	n=read(); m=read();
	for(int i=1; i<=m; i++){
		int u=read(),v=read();
		G[u].push_back(v); G[v].push_back(u);
	}
	if(m&1) {puts("-1"); return 0;}
	dfs(1, -1);
	dfs2(1);
}
### 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>
#define ll long long
#define clr(x,i) memset(x,i,sizeof(x))
using namespace std;
const int N=100005;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
vector<int> G[N];
int n,m,fa[N],dfn[N],f[N],tot;
void dfs(int u,int pre)
{
	fa[u]=pre; dfn[u]=++tot;
	for(auto v:G[u]){
		if(v==pre) continue;
		if(fa[v] && dfn[v]<dfn[u]) f[u]^=1,printf("%d %d\n",u,v);
		else if(!fa[v]) dfs(v, u);
	}
}
void dfs2(int u)
{
	for(auto v:G[u]){
		if(v==fa[u] || fa[v]!=u) continue;
		dfs2(v);
	}
	if(u==1) return;
	if(!f[u]) printf("%d %d\n",fa[u],u),f[fa[u]]^=1;
	else printf("%d %d\n",u,fa[u]),f[u]^=1;
}
int main()
{
	n=read(); m=read();
	for(int i=1; i<=m; i++){
		int u=read(),v=read();
		G[u].push_back(v); G[v].push_back(u);
	}
	if(m&1) {puts("-1"); return 0;}
	dfs(1, -1);
	dfs2(1);
}
```