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

const int N = 1000 * 100 + 100;
int n, m, deg[N], digi[N], h[N];
vector<int> adj[N];
bool vis[N];

void dfs(int root) {
	int v = root;
	vis[root] = true;
	for (auto u: adj[root]) 
		if (!vis[u]) {
			h[u] = h[root] + 1;
			dfs(u);
			if (deg[u] & 1)
				cout << u << ' ' << v << endl, deg[u]++;
			else
				cout << v << ' ' << u << endl, deg[v]++;
		}
		else if (h[root] < h[u]) {
			if (deg[u] & 1)
				cout << u << ' ' << v << endl, deg[u]++;
			else
				cout << v << ' ' << u << endl, deg[root]++;
		}
}

int main() { 
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	for (int i = 1, v, u; i <= m; i++) {
		cin >> v >> u;
		adj[v].push_back(u);
		adj[u].push_back(v);
		digi[v]++, digi[u]++;
	}
	int root = 1;
	for (int i = 1; i <= n; i++)
		if (digi[i] > digi[root])
			root = i;
	if (m & 1)
		return cout << -1 << endl, 0;
	dfs(root);
	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;

const int N = 1000 * 100 + 100;
int n, m, deg[N], digi[N], h[N];
vector<int> adj[N];
bool vis[N];

void dfs(int root) {
	int v = root;
	vis[root] = true;
	for (auto u: adj[root]) 
		if (!vis[u]) {
			h[u] = h[root] + 1;
			dfs(u);
			if (deg[u] & 1)
				cout << u << ' ' << v << endl, deg[u]++;
			else
				cout << v << ' ' << u << endl, deg[v]++;
		}
		else if (h[root] < h[u]) {
			if (deg[u] & 1)
				cout << u << ' ' << v << endl, deg[u]++;
			else
				cout << v << ' ' << u << endl, deg[root]++;
		}
}

int main() { 
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	for (int i = 1, v, u; i <= m; i++) {
		cin >> v >> u;
		adj[v].push_back(u);
		adj[u].push_back(v);
		digi[v]++, digi[u]++;
	}
	int root = 1;
	for (int i = 1; i <= n; i++)
		if (digi[i] > digi[root])
			root = i;
	if (m & 1)
		return cout << -1 << endl, 0;
	dfs(root);
	return 0;
}

```