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

const int MAXN = 100005;

vector <int> G[MAXN];
int N, M, A[MAXN], B[MAXN];
bool vis[MAXN], s[MAXN], x[MAXN];

int fix(int u){
	vis[u] = true;
	
	int r = x[u];
	for(int i : G[u]){
		int v = A[i] ^ B[i] ^ u;
		if(vis[v])continue;
		
		if(fix(v) == 1){
			s[i] ^= 1;
			r ^= 1;
		}
	}
	return r;
}

int main(){		
	scanf("%d %d", &N, &M);
	for(int i = 0; i < M; i++){
		scanf("%d %d", &A[i], &B[i]);
		A[i]--, B[i]--;
		
		G[A[i]].push_back(i);
		G[B[i]].push_back(i);
		
		x[A[i]] ^= 1;
	}
	
	if(M % 2 != 0)printf("-1\n");
	else{
		fix(0);
		for(int i = 0; i < M; i++){
			if(!s[i])printf("%d %d\n", A[i] + 1, B[i] + 1);
			else printf("%d %d\n", B[i] + 1, A[i] + 1);
		}
	}
	
	return 0;
}

### Prompt

In cpp, your task is to solve 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;

const int MAXN = 100005;

vector <int> G[MAXN];
int N, M, A[MAXN], B[MAXN];
bool vis[MAXN], s[MAXN], x[MAXN];

int fix(int u){
	vis[u] = true;
	
	int r = x[u];
	for(int i : G[u]){
		int v = A[i] ^ B[i] ^ u;
		if(vis[v])continue;
		
		if(fix(v) == 1){
			s[i] ^= 1;
			r ^= 1;
		}
	}
	return r;
}

int main(){		
	scanf("%d %d", &N, &M);
	for(int i = 0; i < M; i++){
		scanf("%d %d", &A[i], &B[i]);
		A[i]--, B[i]--;
		
		G[A[i]].push_back(i);
		G[B[i]].push_back(i);
		
		x[A[i]] ^= 1;
	}
	
	if(M % 2 != 0)printf("-1\n");
	else{
		fix(0);
		for(int i = 0; i < M; i++){
			if(!s[i])printf("%d %d\n", A[i] + 1, B[i] + 1);
			else printf("%d %d\n", B[i] + 1, A[i] + 1);
		}
	}
	
	return 0;
}

```