#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < n; i++)
using P = pair<int, int>;
const int mx = 100005;
int n, m, a, b, d[mx];
vector<int> G[mx];
vector<P> ans;
bool dfs(int p, int u){
	bool ptou = true;
	d[u] = d[p]+1;
	for(int nx: G[u]) if(!d[nx]){
		ptou ^= dfs(u,nx);
	}else if(d[u] > d[nx]+1){
		ans.push_back(P(u,nx));
		ptou ^= 1;
	}
	if(p+u) ans.push_back(ptou?P(p,u):P(u,p));
	return ptou;
}
int main(){
	scanf("%d%d", &n, &m);
	if(m%2) return 0*puts("-1");
	rep(i,m){
		scanf("%d%d", &a, &b);
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	dfs(0,0);
	for(P it: ans) printf("%d %d\n", it.first+1, it.second+1);
}
### Prompt

Your challenge is to write 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 rep(i,n) for(int i = 0; i < n; i++)
using P = pair<int, int>;
const int mx = 100005;
int n, m, a, b, d[mx];
vector<int> G[mx];
vector<P> ans;
bool dfs(int p, int u){
	bool ptou = true;
	d[u] = d[p]+1;
	for(int nx: G[u]) if(!d[nx]){
		ptou ^= dfs(u,nx);
	}else if(d[u] > d[nx]+1){
		ans.push_back(P(u,nx));
		ptou ^= 1;
	}
	if(p+u) ans.push_back(ptou?P(p,u):P(u,p));
	return ptou;
}
int main(){
	scanf("%d%d", &n, &m);
	if(m%2) return 0*puts("-1");
	rep(i,m){
		scanf("%d%d", &a, &b);
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	dfs(0,0);
	for(P it: ans) printf("%d %d\n", it.first+1, it.second+1);
}
```