#include <bits/stdc++.h>
using namespace std;
using PII = pair<int, int>;
using LL = long long;
vector<int> G[100005];
int deg[100005], depth[100005];
vector<PII> edges;
void dfs(int from, int id) {
	depth[id] = depth[from] + 1;
	int x = 0;
	for (auto x: G[id]) {
		if (depth[x] == -1) {
			dfs(id, x);
		} else {
			if (depth[x] > depth[id]) {
				edges.emplace_back(id, x);
				deg[id]++;
			}
		}
	}
	if (deg[id] % 2) {
		if (from != 0) {
			edges.emplace_back(id, from);
			deg[id]++;
		}
	} else {
		if (from != 0) {
			edges.emplace_back(from, id);
			deg[from]++;
		}
	}
}

int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0; i <= n; i++) depth[i] = -1;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	if (m % 2 == 1) {
		cout << "-1\n";
		return 0;
	}
	dfs(0, 1);
	for (auto p: edges) {
		cout << p.first << ' ' << p.second << '\n';
	}
	return 0;
}

### Prompt

Your task is to create 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;
using PII = pair<int, int>;
using LL = long long;
vector<int> G[100005];
int deg[100005], depth[100005];
vector<PII> edges;
void dfs(int from, int id) {
	depth[id] = depth[from] + 1;
	int x = 0;
	for (auto x: G[id]) {
		if (depth[x] == -1) {
			dfs(id, x);
		} else {
			if (depth[x] > depth[id]) {
				edges.emplace_back(id, x);
				deg[id]++;
			}
		}
	}
	if (deg[id] % 2) {
		if (from != 0) {
			edges.emplace_back(id, from);
			deg[id]++;
		}
	} else {
		if (from != 0) {
			edges.emplace_back(from, id);
			deg[from]++;
		}
	}
}

int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0; i <= n; i++) depth[i] = -1;
	for (int i = 0; i < m; i++) {
		int a, b;
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	if (m % 2 == 1) {
		cout << "-1\n";
		return 0;
	}
	dfs(0, 1);
	for (auto p: edges) {
		cout << p.first << ' ' << p.second << '\n';
	}
	return 0;
}

```