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 maxn = 100001;
vector<int> adj[maxn];
int dep[maxn], deg[maxn];

void dfs(int u, int d) {
	if (dep[u]) return;
	dep[u] = d;
	vector<int> anc;
	for (const auto& v : adj[u]) dfs(v, d + 1);
	for (const auto& v : adj[u]) if (dep[v] < dep[u]) {
		if (deg[u]) {
			cout << u << ' ' << v << '\n';
			deg[u] = 0;
		} else {
			cout << v << ' ' << u << '\n';
			deg[v] ^= 1;
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	int N, M;
	cin >> N >> M;
	for (int i = 0; i != M; ++i) {
		int u, v;
		cin >> u >> v;
		adj[u].emplace_back(v);
		adj[v].emplace_back(u);
	}
	if (M & 1) return cout << -1 << endl, 0;
	dfs(1, 1);
}