#include <bits/stdc++.h>
using namespace std;
						
#define fast_cin() ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define endl '\n'
#define int long long

const int N = 1e5 + 5;

int a[N], b[N], out_degree[N];
vector<pair<int, int>> g[N];
vector<bool> seen(N, false);

void dfs(int u, int id);
				
int32_t main() {
	fast_cin();  
	int n, m; cin >> n >> m;
	if (m & 1) {
		cout << -1 << endl;
		return 0;
	}
	for (int i = 1; i <= m; ++i) {
		cin >> a[i] >> b[i];
		g[a[i]].push_back({b[i], i});
		g[b[i]].push_back({a[i], i});
		++out_degree[a[i]];
	}
	dfs(1, 0);
	for (int i = 1; i <= m; ++i) {
		cout << a[i] << " " << b[i] << endl;
	}
}

void dfs(int u, int id) {
	seen[u] = true;
	for (auto &e : g[u]) {
		auto [v, n_id] = e;
		if (!seen[v]) {
			dfs(v, n_id);
		}
	}
	if (out_degree[u] & 1) {
		assert(id != 0);
		--out_degree[a[id]], ++out_degree[b[id]];
		swap(a[id], b[id]);
	}
}
### Prompt

Generate 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;
						
#define fast_cin() ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define endl '\n'
#define int long long

const int N = 1e5 + 5;

int a[N], b[N], out_degree[N];
vector<pair<int, int>> g[N];
vector<bool> seen(N, false);

void dfs(int u, int id);
				
int32_t main() {
	fast_cin();  
	int n, m; cin >> n >> m;
	if (m & 1) {
		cout << -1 << endl;
		return 0;
	}
	for (int i = 1; i <= m; ++i) {
		cin >> a[i] >> b[i];
		g[a[i]].push_back({b[i], i});
		g[b[i]].push_back({a[i], i});
		++out_degree[a[i]];
	}
	dfs(1, 0);
	for (int i = 1; i <= m; ++i) {
		cout << a[i] << " " << b[i] << endl;
	}
}

void dfs(int u, int id) {
	seen[u] = true;
	for (auto &e : g[u]) {
		auto [v, n_id] = e;
		if (!seen[v]) {
			dfs(v, n_id);
		}
	}
	if (out_degree[u] & 1) {
		assert(id != 0);
		--out_degree[a[id]], ++out_degree[b[id]];
		swap(a[id], b[id]);
	}
}
```