#include <iostream>
#include <cstdio>
#include <vector>
#define N 100005
using namespace std;

int n, m, a, b, c[N], d[N];
vector<int> gr[N];

void f(int p, int q) {
	d[p] = d[q] + 1;
	for(auto x:gr[p]) {
		if(x == q) continue;
		if(d[x]) {
			if(d[x] < d[p]) {
				c[p] ^= 1;
				printf("%d %d\n", p, x);
			}
		} else {
			f(x, p);
		}
	}
	if(c[p]) printf("%d %d\n", p, q);
	else if(q) {
		c[q] ^= 1;
		printf("%d %d\n", q, p);
	}
}

int main() {
	cin >> n >> m;
	if(m&1) return puts("-1")*0;
	while(m--) {
		scanf("%d%d", &a, &b);
		gr[a].push_back(b);
		gr[b].push_back(a);
	}
	f(1, 0);
	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 <iostream>
#include <cstdio>
#include <vector>
#define N 100005
using namespace std;

int n, m, a, b, c[N], d[N];
vector<int> gr[N];

void f(int p, int q) {
	d[p] = d[q] + 1;
	for(auto x:gr[p]) {
		if(x == q) continue;
		if(d[x]) {
			if(d[x] < d[p]) {
				c[p] ^= 1;
				printf("%d %d\n", p, x);
			}
		} else {
			f(x, p);
		}
	}
	if(c[p]) printf("%d %d\n", p, q);
	else if(q) {
		c[q] ^= 1;
		printf("%d %d\n", q, p);
	}
}

int main() {
	cin >> n >> m;
	if(m&1) return puts("-1")*0;
	while(m--) {
		scanf("%d%d", &a, &b);
		gr[a].push_back(b);
		gr[b].push_back(a);
	}
	f(1, 0);
	return 0;
}
```