#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+5;
int u[N], v[N], n, m;

int hed[N],to[N<<1],nxt[N<<1],cnt;
inline void adde(int u,int v){
	++cnt;to[cnt]=v,nxt[cnt]=hed[u];hed[u]=cnt;
}

int col[N],vis[N],dfn[N],num;

inline void dfs(int x,int pre){
	vis[x]=1,dfn[x]=++num;
	for(int i=hed[x];i;i=nxt[i]){
		int v=to[i];
		if(!vis[v])dfs(v,x);
	}
	for(int i=hed[x];i;i=nxt[i]){
		int v=to[i];
		if(dfn[v]>dfn[x]){
			if(col[v]){
				printf("%d %d\n",v,x);
				col[v]^=1;
			}else{
				printf("%d %d\n",x,v);
				col[x]^=1;
			}
		}
	}
}

int main()
{
	cin >> n >> m;
	if(m%2){puts("-1");return 0;}
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u[i],&v[i]);
		adde(u[i],v[i]);
		adde(v[i],u[i]);
	}
	dfs(1,0);
}

### 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+5;
int u[N], v[N], n, m;

int hed[N],to[N<<1],nxt[N<<1],cnt;
inline void adde(int u,int v){
	++cnt;to[cnt]=v,nxt[cnt]=hed[u];hed[u]=cnt;
}

int col[N],vis[N],dfn[N],num;

inline void dfs(int x,int pre){
	vis[x]=1,dfn[x]=++num;
	for(int i=hed[x];i;i=nxt[i]){
		int v=to[i];
		if(!vis[v])dfs(v,x);
	}
	for(int i=hed[x];i;i=nxt[i]){
		int v=to[i];
		if(dfn[v]>dfn[x]){
			if(col[v]){
				printf("%d %d\n",v,x);
				col[v]^=1;
			}else{
				printf("%d %d\n",x,v);
				col[x]^=1;
			}
		}
	}
}

int main()
{
	cin >> n >> m;
	if(m%2){puts("-1");return 0;}
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u[i],&v[i]);
		adde(u[i],v[i]);
		adde(v[i],u[i]);
	}
	dfs(1,0);
}

```