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 ll = long long;
using namespace std;

const int N = 1e5 + 1, mod = 1e9 + 7;

int n, m, total = 0, h[N], deg[N];
bool nc[N];
vector<int> adj[N];
vector<pair<int, int>> ans;

void dfs(int v, int p) {
	for (auto u : adj[v]) {
		if (u == p) {
			continue;
		}
		if (h[u] == 0) {
			h[u] = h[v] + 1;
			dfs(u, v);
		}
		if (!nc[u] && (h[u] == h[v] + 1 || (h[u] && h[u] < h[v]))) {
			deg[v]++;
			ans.push_back({v, u});
		}
	}
	if (h[v] > 1 && (deg[v] & 1)) {
		deg[v]++;
		ans.push_back({v, p});
		nc[v] = 1;
	}
}

int main() { cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	for (int i = 1, u, v; i <= m; i++) {
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	h[1] = 1;
	dfs(1, 0);
	for (int v = 1; v <= n; v++) {
		if (deg[v] % 2 == 0) {
			total++;
		}
	}
	if (total != n) {
		cout << -1;
		return 0;
	}
	for (auto cur : ans) {
		cout << cur.first << ' ' << cur.second << '\n';
	}
}