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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int n,m,cc,u[N],v[N],hd[N],vv[N],d[N],vis[N];
struct E{int v,nxt;}e[N<<1];
void add(int u,int v){e[++cc]=(E){v,hd[u]};hd[u]=cc;e[++cc]=(E){u,hd[v]};hd[v]=cc;}
void dfs(int x)
{
	vv[x]=1;
	for(int i=hd[x];i;i=e[i].nxt)
	{
		int y=e[i].v;if(vv[y])continue;
		vis[i>>1]=1;dfs(y);
	}
}
void dfs2(int x)
{
	vv[x]=0;
	for(int i=hd[x];i;i=e[i].nxt)
	{
		int y=e[i].v;if(!vv[y])continue;
		dfs2(y);if(!d[y])vis[i>>1]=i&1,d[x]^=1;else vis[i>>1]=(i&1)^1,d[y]^=1;
	}
}
int main()
{
	scanf("%d%d",&n,&m);cc=1;
	if(m&1){puts("-1");return 0;}
	for(int i=1;i<=m;i++){scanf("%d%d",&u[i],&v[i]);add(u[i],v[i]);}
	dfs(1);for(int i=1;i<=m;i++)if(!vis[i])d[u[i]]^=1;dfs2(1);
	for(int i=1;i<=m;i++)if(!vis[i])printf("%d %d\n",u[i],v[i]);else printf("%d %d\n",v[i],u[i]);
	return 0;
}