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 _ = 1e5 + 7;
struct Edge{int end , upEd;}Ed[_ << 1];
int head[_] , dir[_] , d[_] , dep[_] , cntEd = 1 , N , M;

void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd;}

int dfs(int x , int p){
	dep[x] = dep[p] + 1;
	for(int i = head[x] ; i ; i = Ed[i].upEd)
		if(Ed[i].end != p)
			if(!dep[Ed[i].end]){int t = dfs(Ed[i].end , x); dir[i >> 1] = (i & 1) ^ t; if(!t) ++d[x];}
			else if(dep[Ed[i].end] < dep[x]){++d[x]; dir[i >> 1] = i & 1;}
	return d[x] & 1;
}

int main(){
	cin >> N >> M; for(int i = 1 ; i <= M ; ++i){int p , q; cin >> p >> q; addEd(q , p); addEd(p , q);}
	if(M & 1){puts("-1"); return 0;}
	dfs(1 , 0);
	for(int i = 1 ; i <= M ; ++i)
		printf("%d %d\n" , Ed[(i << 1 | dir[i]) ^ 1].end , Ed[i << 1 | dir[i]].end);
	return 0;
}