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;

struct Edge {
  int t,next;
  Edge() {}
  Edge(int a,int b):t(a),next(b) {}
};

Edge e[200005];
int head[100005];
bool col[100005],d[100005];

bool vis[100005];

bool dfs(int x) {
  vis[x]=1;
  bool s=d[x];
  for(int i=head[x];i;i=e[i].next)
    if (!vis[e[i].t]) {
    	bool v=dfs(e[i].t);
    	s^=v;
    	if (v) col[(i+1)>>1]^=1;
	}
  return s;
}

int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  if (m&1) {
  	puts("-1");
  	return 0;
  }
  for(int i=1;i<=m;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	e[2*i-1]=Edge(y,head[x]);
  	head[x]=2*i-1;
  	e[2*i]=Edge(x,head[y]);
  	head[y]=2*i;
  	d[x]^=1;
  }
  dfs(1);
  for(int i=1;i<=m;i++)
    if (!col[i]) printf("%d %d\n",e[2*i].t,e[2*i-1].t);
    else printf("%d %d\n",e[2*i-1].t,e[2*i].t);
  return 0;
}