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

typedef long long ll;
const int N = 300100;

struct edge{
	int ne, to;
}e[N];
int he[N], du[N], vis[N], tot = 0, n, m, ban[N], exist[N];

void add(int x, int y){
	e[++tot].ne = he[x];
	he[x] = tot;
	e[tot].to = y;
}

void dfs(int x, int fa){
	int i, y;
	vis[x] = 1;
	exist[x] = 1;
	for (i = he[x]; i; i = e[i].ne){
		y = e[i].to;
		if (y != fa){
			if (!vis[y]){
				dfs(y, x);
			}
			else if (exist[y]){
				printf("%d %d\n", x, y);
				du[x] ^= 1;
			}
		}
	}
	if (x == 1){
		return;
	}
	if (du[x]){
		printf("%d %d\n", x, fa);
		du[x] ^= 1;
	}
	else{
		printf("%d %d\n", fa, x);
		du[fa] ^= 1;
	}
	exist[x] = 0;
}

int main(){
	//freopen("a.in", "r", stdin);
	//freopen("a.out", "w", stdout);
	int i, j, k, x, y;
	scanf("%d%d", &n, &m);
	if (m % 2 != 0){
		printf("-1\n");
		return 0;
	}
	for (i = 1; i <= m; i++){
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	return 0;
}
### 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;

typedef long long ll;
const int N = 300100;

struct edge{
	int ne, to;
}e[N];
int he[N], du[N], vis[N], tot = 0, n, m, ban[N], exist[N];

void add(int x, int y){
	e[++tot].ne = he[x];
	he[x] = tot;
	e[tot].to = y;
}

void dfs(int x, int fa){
	int i, y;
	vis[x] = 1;
	exist[x] = 1;
	for (i = he[x]; i; i = e[i].ne){
		y = e[i].to;
		if (y != fa){
			if (!vis[y]){
				dfs(y, x);
			}
			else if (exist[y]){
				printf("%d %d\n", x, y);
				du[x] ^= 1;
			}
		}
	}
	if (x == 1){
		return;
	}
	if (du[x]){
		printf("%d %d\n", x, fa);
		du[x] ^= 1;
	}
	else{
		printf("%d %d\n", fa, x);
		du[fa] ^= 1;
	}
	exist[x] = 0;
}

int main(){
	//freopen("a.in", "r", stdin);
	//freopen("a.out", "w", stdout);
	int i, j, k, x, y;
	scanf("%d%d", &n, &m);
	if (m % 2 != 0){
		printf("-1\n");
		return 0;
	}
	for (i = 1; i <= m; i++){
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	return 0;
}
```