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;
const int MAXA=3e5;
int nxt[MAXA],first[MAXA],go[MAXA],tot,tott;
int bo[MAXA];
int add(int x,int y){
    nxt[++tot]=first[x]; first[x]=tot; go[tot]=y;
}
int dfs(int x,int father){
    int tot=0,y;
    bo[x]=++tott;
    for(int i=first[x];i;i=nxt[i])
	if(go[i]!=father){
	    if(!bo[go[i]]){
		y=dfs(go[i],x);
		if(y) printf("%d %d\n",x,go[i]);
		else printf("%d %d\n",go[i],x);
		tot+=y;
	    }else{
		if(bo[go[i]]<bo[x]) continue;
		printf("%d %d\n",x,go[i]);
		tot++;
	    }
	}
    if(tot%2) return 0;
    else return 1;
}

int main(){
    int n,m,x,y;
    scanf("%d%d",&n,&m);
    if(m&1) return printf("-1"),0;
    for(int i=1;i<=m;i++){
	scanf("%d%d",&x,&y);
	add(x,y);
	add(y,x);
    }
    dfs(1,0);
}