#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second

struct edge
{
	int u, v, id;
};

int main()
{
	#ifdef DGC
		//freopen("a.txt", "r", stdin);
		//freopen("b.txt", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	int n, m;
	cin >> n >> m;
	vector<bool> deg(n);
	vector<vector<edge>> adj(n);
	vector<pair<int, int>> how(m);
	for (int i = 0, u, v; i < m; ++i)
	{
		cin >> u >> v;
		--u, --v;
		adj[u].push_back({ u, v, i });
		adj[v].push_back({ v, u, i });
		how[i] = { u, v };
		deg[u] = deg[u]^true;
	}

	vector<bool> mark(n);
	function<void(int, int, int)> dfs = [&](int u, int p, int id)
	{
		mark[u] = true;
		for (auto v : adj[u])
			if (!mark[v.v])
				dfs(v.v, u, v.id);
		if (deg[u] && p != -1)
		{
			swap(how[id].F, how[id].S);
			deg[p] = deg[p]^true;
		}
	};

	dfs(0, -1, -1);
	if (!deg[0])
	{
		for (auto i : how)
			cout << i.F+1 << " " << i.S+1 << "\n";
	}
	else
		cout << "-1\n";

	return 0;
}

### Prompt

In Cpp, your task is to solve 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;

typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second

struct edge
{
	int u, v, id;
};

int main()
{
	#ifdef DGC
		//freopen("a.txt", "r", stdin);
		//freopen("b.txt", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	int n, m;
	cin >> n >> m;
	vector<bool> deg(n);
	vector<vector<edge>> adj(n);
	vector<pair<int, int>> how(m);
	for (int i = 0, u, v; i < m; ++i)
	{
		cin >> u >> v;
		--u, --v;
		adj[u].push_back({ u, v, i });
		adj[v].push_back({ v, u, i });
		how[i] = { u, v };
		deg[u] = deg[u]^true;
	}

	vector<bool> mark(n);
	function<void(int, int, int)> dfs = [&](int u, int p, int id)
	{
		mark[u] = true;
		for (auto v : adj[u])
			if (!mark[v.v])
				dfs(v.v, u, v.id);
		if (deg[u] && p != -1)
		{
			swap(how[id].F, how[id].S);
			deg[p] = deg[p]^true;
		}
	};

	dfs(0, -1, -1);
	if (!deg[0])
	{
		for (auto i : how)
			cout << i.F+1 << " " << i.S+1 << "\n";
	}
	else
		cout << "-1\n";

	return 0;
}

```