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;
typedef long long ll;
typedef unsigned long long hxtype;
typedef pair<int,int>pii;
#define rep(i,l,r) for(i=(l);i<=(r);++i)
#define per(i,l,r) for(i=(l);i>=(r);--i)
#define REP(i,l,r) for(i=(l);i< (r);++i)
#define PER(i,l,r) for(i=(l);i> (r);--i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
template<class IT>inline void cmax(IT &a,IT b){if(a<b)a=b;}
template<class IT>inline void cmin(IT &a,IT b){if(b<a)a=b;}
const ll p=998244353ll;
const int N=100005;
vector<pii>e[N];
pii ed[N];
int d[N],vis[N];
void dfs(int v){
	vis[v]=1;
	int s=e[v].size(),u,dv=d[v],w;
	while(s--)if(!vis[u=e[v][s].fi]){
		dfs(u);
		if(d[u]){
			dv^=1;
			w=e[v][s].se;
			swap(ed[w].fi,ed[w].se);
		}
	}
	d[v]=dv;
}
int main(){
	int n,m,i,v,u;
	scanf("%d%d",&n,&m);
	rep(i,1,m){
		scanf("%d%d",&v,&u);
		d[v]^=1;
		e[v].push_back(mp(u,i));
		e[u].push_back(mp(v,i));
		ed[i]=mp(v,u);
	}
	rep(i,1,n)if(!vis[i]){
		dfs(i);
		if(d[i]){putchar('-');putchar('1');return 0;}
	}
	rep(i,1,m)printf("%d %d\n",ed[i].fi,ed[i].se);
	return 0;
}