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>

#ifdef NON_SUBMIT
#define TEST(n) (n)
#define tout cerr
#else
#define TEST(n) ((void)0)
#define tout cin
#endif

using namespace std;

vector<int> adj[100000];
vector<pair<int,int>> ans;
int cnt[100000], parent[100000];
bool chk[100000];

void dfs(int c)
{
	chk[c]=true;
	for(auto n: adj[c]) {
		if(parent[n]==-1) {
			parent[n]=c;
			dfs(n);
		}
		else if(chk[n] && parent[c]!=n) {
			cnt[c]=1-cnt[c];
			ans.emplace_back(c,n);
		}
	}
	if(c) {
		if(cnt[c]) ans.emplace_back(c,parent[c]);
		else {
			cnt[parent[c]]=1-cnt[parent[c]];
			ans.emplace_back(parent[c],c);
		}
	}
	chk[c]=false;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	TEST(freopen("input.txt","r",stdin));
	TEST(freopen("output.txt","w",stdout));
	TEST(freopen("debug.txt","w",stderr));
	int N, M;
	for(cin>>N>>M;M--;) {
		int a, b;
		cin>>a>>b;
		adj[--a].push_back(--b);
		adj[b].push_back(a);
	}
	memset(parent,-1,sizeof(parent));
	parent[0]=0;
	dfs(0);
	if(cnt[0]) cout<<"-1\n";
	else {
		for(auto e: ans) cout<<e.first+1<<' '<<e.second+1<<'\n';
	}
	return 0;
}