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;

#define N 100010
int n, m;
vector<int> g[N];
int rt[N];

int findrt(int x) {
	return rt[x] = (rt[x] == x ? x:findrt(rt[x]));
}

int deg[N];

void link(int a, int b) {
	cout<<a<<' '<<b<<'\n';
	deg[a] ^= 1;
}

void dfs(int u, int pa) {
	for (int v:g[u]) if (v != pa) {
		dfs(v, u);
		if (deg[v]) link(v, u);
		else link(u, v);
	}
}

int main() {
//35b
	cin>>n>>m;
	if (m & 1) {
		cout<<-1;
		return 0;
	}
	for (int i = 1; i <= n; i ++) rt[i] = i;
	while (m --) {
		int a, b;
		cin>>a>>b;
		int x = findrt(a), y = findrt(b);
		if (x == y) link(a, b);
		else {
			g[a].push_back(b);
			g[b].push_back(a);
			rt[x] = y;
		}
	}
	dfs(1, 1);
}