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 N = 1 * 100 * 1000 + 1;
long long n, m, dp[N], h[N];
bool vis[N];
vector <int> adj[N], ans[N];

void dfs(int u) {
	vis[u] = true;
	for (int v: adj[u]) {
		int cnt = 0;
		if (vis[v] == false) {
			cnt++;
			h[v] = h[u] + 1;
			dfs(v);
			dp[v] % 2 == 1? ans[v].push_back(u): ans[u].push_back(v);
			dp[u] += dp[v];
		}
		else if (h[v] > h[u]) {
			cnt++;
			ans[u].push_back(v);
		}
		dp[u] += cnt;
	}
}

int main() {
	
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	
	cin >> n >> m;
	for (int i = 0, u, v; i < m; i++) {
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	if (m % 2 == 1)
		return cout << "-1\n", 0;
	dfs(1);
	for (int i = 1; i <= n; i++)
		for (int v: ans[i])
			cout << i << ' ' << v << endl;
	
	return 0;
}