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>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=100005;

int n,m,Dfn;
vector<int>G[N];
vector<pair<int,int>>edge1,edge2;
int du[N],dfn[N],dep[N];
bool vis[N],defeat=false;

void Dfs(int u,int fa) {
	vis[u]=true;
	dfn[u]=Dfn++;
	dep[u]=dep[fa]+1;
	for(int i=0;i<G[u].size();i++) {
		int v=G[u][i];
		if(vis[v]) {
			du[v]++;
			if(v!=fa&&dep[u]>dep[v])
				edge1.push_back(make_pair(v,u));
			continue;
		}
		edge2.push_back(make_pair(u,v));
		Dfs(v,u);
	}
	if(du[u]%2==1) {
		if(u==1) {defeat=true;return ;}
		du[u]++,du[fa]--;
		edge2[dfn[u]-1]=make_pair(u,fa);
	}
}
int main() {
	scanf("%d %d",&n,&m);
	for(int i=1,u,v;i<=m;i++) {
		scanf("%d %d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	Dfs(1,1);
	if(defeat) {
		printf("-1\n");
		return 0;
	}
	for(int i=0;i<edge1.size();i++)
		printf("%d %d\n",edge1[i].first,edge1[i].second);
	for(int i=0;i<edge2.size();i++)
		printf("%d %d\n",edge2[i].first,edge2[i].second);
	return 0;
}