#include <bits/stdc++.h>
using namespace std;

vector<vector<int>> g;
vector<pair<int, int>> e;
int t[111111], d[111111];
void f(int u, int x) {
	if (t[x]) return;
	t[x] = 1;
	int v = u ? u ^ e[x].first^e[x].second : 1;
	for (int y : g[v])
		if (y != x) f(v, y);
	if (!u) return;
	if (d[v]) cout << v << ' ' << u << '\n', d[v] = 0;
	else cout << u << ' ' << v << '\n', d[u] ^= 1;
}
int main() {
	int n, m;
	cin >> n >> m;
	if (m & 1) {
		cout << -1 << endl;
		return 0;
	}
	g.resize(n + 1);
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		e.emplace_back(a, b);
		g[a].push_back(i);
		g[b].push_back(i);
	}
	f(0, m);
	cout << flush;
	return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

vector<vector<int>> g;
vector<pair<int, int>> e;
int t[111111], d[111111];
void f(int u, int x) {
	if (t[x]) return;
	t[x] = 1;
	int v = u ? u ^ e[x].first^e[x].second : 1;
	for (int y : g[v])
		if (y != x) f(v, y);
	if (!u) return;
	if (d[v]) cout << v << ' ' << u << '\n', d[v] = 0;
	else cout << u << ' ' << v << '\n', d[u] ^= 1;
}
int main() {
	int n, m;
	cin >> n >> m;
	if (m & 1) {
		cout << -1 << endl;
		return 0;
	}
	g.resize(n + 1);
	for (int i = 0; i < m; i++) {
		int a, b;
		scanf("%d%d", &a, &b);
		e.emplace_back(a, b);
		g[a].push_back(i);
		g[b].push_back(i);
	}
	f(0, m);
	cout << flush;
	return 0;
}

```