#include <bits/stdc++.h>

using namespace std;

const int MX = 100000;

vector<int> G[MX];
int vis[MX];

int dfs(int v, int p = -1) {
	vis[v] = 1;
	int cnt = 0;
	for (int u : G[v]) {
		if (u == p) continue;
		if (vis[u] == 1) {
			printf("%d %d\n", v + 1, u + 1);
			cnt ^= 1;
		}
		else if (vis[u] == 0) {
			cnt ^= dfs(u, v);
		}
	}
	
	vis[v] = 2;
	
	if (cnt == 0) {
		if (p != -1) printf("%d %d\n", p + 1, v + 1);
		return 1;
	}
	else {
		if (p != -1) printf("%d %d\n", v + 1, p + 1);
		return 0;
	}
}

int main() {
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	
	if (m % 2 == 1) {
		printf("%d\n", -1);
		return 0;
	}
	
	for (int i = 0, u, v; i < m; i++) {
		ignore = scanf("%d %d", &u, &v);
		u--; v--;
		G[u].emplace_back(v);
		G[v].emplace_back(u);
	}
	
	dfs(0);
	
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 MX = 100000;

vector<int> G[MX];
int vis[MX];

int dfs(int v, int p = -1) {
	vis[v] = 1;
	int cnt = 0;
	for (int u : G[v]) {
		if (u == p) continue;
		if (vis[u] == 1) {
			printf("%d %d\n", v + 1, u + 1);
			cnt ^= 1;
		}
		else if (vis[u] == 0) {
			cnt ^= dfs(u, v);
		}
	}
	
	vis[v] = 2;
	
	if (cnt == 0) {
		if (p != -1) printf("%d %d\n", p + 1, v + 1);
		return 1;
	}
	else {
		if (p != -1) printf("%d %d\n", v + 1, p + 1);
		return 0;
	}
}

int main() {
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	
	if (m % 2 == 1) {
		printf("%d\n", -1);
		return 0;
	}
	
	for (int i = 0, u, v; i < m; i++) {
		ignore = scanf("%d %d", &u, &v);
		u--; v--;
		G[u].emplace_back(v);
		G[v].emplace_back(u);
	}
	
	dfs(0);
	
	return 0;
}

```