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<cstdio>
#include<vector>
using namespace std;

vector<int>g[100001];
bool vis[100001],vis2[100001],w[100001];
struct ed{int u,v;};
ed ans[100001];
int anst;
void dfs(int u){
	vis[u]=1;
	for(int i=0;i<g[u].size();i++)
		if(!vis[g[u][i]])
			dfs(g[u][i]);
	for(int i=0;i<g[u].size();i++)
		if(!vis2[g[u][i]]){
			w[g[u][i]]^=1;
			ans[++anst]=(ed){g[u][i],u};
		}
	if(w[u]){
		swap(ans[anst].u,ans[anst].v);
		w[ans[anst].u]^=1;
		w[ans[anst].v]^=1;
	}
	vis2[u]=1;
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	if(m&1){
		printf("-1\n");
		return 0;
	}
	dfs(1);
	for(int i=1;i<=anst;i++)
		printf("%d %d\n",ans[i].u,ans[i].v);
}