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>
#define fo(i, x, y) for(int i = x; i <= y; i ++)
using namespace std;

const int N = 5e5 + 5;

int T,n,m,x,y;
int fi[N],to[N * 2],nt[N * 2],tot,bz[N];
int fa[N],ans[N],siz[N];

void link(int x,int y){
	nt[++tot] = fi[x],to[tot] = y,fi[x] = tot;
}

void cl(){
	fo(i,1,n)	bz[i] = fi[i] = ans[i] = siz[i] = 0;
	tot = 1;
}

void dg(int x,int la){
	bz[x] = 1;
	for(int i = fi[x]; i; i = nt[i]) if(!bz[to[i]]){
		ans[i / 2] = (i & 1);
		siz[to[i]] ^= 1;
		dg(to[i],i);
		siz[x] ^= siz[to[i]];
	} else if(i != (la ^ 1)){
		ans[i / 2] = !(i & 1);
		siz[to[i]] ^= 1;
	}
	if(siz[x]) ans[la / 2] ^= 1;
}

#include "vector";
using namespace std;

int main(){
	//for(scanf("%d",&T); T; T--){
		cl();
		scanf("%d %d",&n,&m);
		vector<pair<int,int>>ed;
		fo(i,1,m){
			scanf("%d %d",&x,&y);
			ed.push_back({x,y});
			link(x,y); link(y,x);
		}
		if(m & 1){
			printf("-1\n"); return 0;
		}
		dg(1,0);
		fo(i,1,m){
			printf("%d %d\n",(ans[i]?ed[i-1].first:ed[i-1].second)
				   ,(ans[i]?ed[i-1].second:ed[i-1].first));
		}
	//}
}

//sub-EOF