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 mp make_pair
#define fi first
#define se second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n,m;
int f[100005],du[100005];
vector<int> adj[100005];
bool vis[100005];

void dfs1(int u){
	for(auto x:adj[u]){
		if(vis[x]) continue;
		vis[x]=true;
		f[x]=u;
		dfs1(x);
	}
}

void dfs2(int u){
	for(auto x:adj[u]){
		if(f[x]!=u) continue;
		dfs2(x);
		if(du[x]==1) printf("%d %d\n",x,u);
		else printf("%d %d\n",u,x),du[u]^=1;
	}
}

int main(){
	n=readint(); m=readint();
	int x,y;
	for(int i=1;i<=m;i++){
		x=readint(); y=readint();
		adj[y].push_back(x),adj[x].push_back(y);
	}
	if(m%2==1) return printf("-1\n"),0;
	vis[1]=true; dfs1(1);
	for(int i=1;i<=n;i++) for(auto x:adj[i]) if(i<=x&&f[x]!=i&&f[i]!=x) printf("%d %d\n",i,x),du[i]^=1;
	dfs2(1);
	return 0;
}