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>

#define f first
#define s second
#define pb push_back
#define mp make_pair

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;

const int N = (int)2e5 + 123, inf = 1e9, mod = 1e9 + 7;
const ll INF = 1e18;

vector< int > g[N];
int n, m, s[N], t[N];
bool used[N], deg[N];
map<int, int> e[N];
void dfs(int v, int pr = -1){
	used[v] = 1;
	for(auto to : g[v]){
		if(pr == to)
			continue;
		if(used[to]){
			if(e[to][v] == 0 && e[v][to] == 0){
				e[v][to] = 1;
				deg[v] ^= 1;
			}
		}else{
			dfs(to, v);
			if(deg[to])
				e[to][v] = 1, deg[to] ^= 1;
			else
				e[v][to] = 1, deg[v] ^= 1;
		}	
	}
}
int main()
{
	scanf("%d%d", &n, &m);
	if(m & 1){
		printf("-1");
		return 0;
	}

	for(int i = 0; i < m; i++){
		scanf("%d%d", &s[i], &t[i]);
		g[s[i]].pb(t[i]);
		g[t[i]].pb(s[i]);
	}
	dfs(1);
	for(int i = 0; i < m; i++){
		if(e[s[i]][t[i]] == 0)
			swap(s[i], t[i]);			
		printf("%d %d\n", s[i], t[i]);
	}
		
	return 0;
}