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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

const int maxn=1e5+5;

int n,m;

int tot,go[2*maxn],id[2*maxn],nxt[2*maxn],f1[maxn];
void ins(int x,int y,int i)
{
	go[++tot]=y;
	id[tot]=i;
	nxt[tot]=f1[x];
	f1[x]=tot;
}

bool vis[maxn],dfn[maxn],dg[maxn];
int e[maxn][2],e0;
void dfs(int k,int last)
{
	dfn[k]=1;
	for(int p=f1[k]; p; p=nxt[p]) if (!vis[id[p]])
	{
		vis[id[p]]=1;
		if (!dfn[go[p]])
		{
			dfs(go[p],k);
		} else
		{
			e[++e0][0]=k, e[e0][1]=go[p];
			dg[k]^=1;
		}
	}
	if (!last) return;
	if (dg[k]) e[++e0][0]=k, e[e0][1]=last;
		else e[++e0][0]=last, e[e0][1]=k, dg[last]^=1;
}

int main()
{
	scanf("%d %d",&n,&m);
	fo(i,1,m)
	{
		int x,y;
		scanf("%d %d",&x,&y);
		ins(x,y,i), ins(y,x,i);
	}
	
	dfs(1,0);
	
	if (dg[1]) puts("-1"); else
	{
		fo(i,1,m) printf("%d %d\n",e[i][0],e[i][1]);
	}
}