#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int> ii;
const int N = 1e5 + 5;
int n, m;
int out[N], cnt[N];
vector<ii> G[N];
ii ans[N];
bool vis[N];
#define fi first
#define se second
void dfs(int u){
	vis[u] = 1;
	cnt[u] = out[u];
	for(auto x : G[u]){
		int v = x.fi;
		if(!vis[v]){
			dfs(v);
			cnt[u]+=cnt[v];
			cnt[u]%=2;
			if(cnt[v]%2) swap(ans[x.se].fi,ans[x.se].se); 
		}
	}
}
int main(){
	cin >> n >> m;
	for(int i=1;i<=m;i++){
		int u, v;
		cin >> u >> v;
		out[u]++;
		G[u].push_back({v,i});
		G[v].push_back({u,i});
		ans[i] = {u,v};
	}
	int tmp = 0;
	for(int i=1;i<=n;i++) out[i] %= 2, tmp+= out[i];
	if(tmp%2) return cout << -1, 0;
	dfs(1);
	for(int i=1;i<=m;i++) cout << ans[i].fi << " " << ans[i].se << endl;
}
### 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 pair<int,int> ii;
const int N = 1e5 + 5;
int n, m;
int out[N], cnt[N];
vector<ii> G[N];
ii ans[N];
bool vis[N];
#define fi first
#define se second
void dfs(int u){
	vis[u] = 1;
	cnt[u] = out[u];
	for(auto x : G[u]){
		int v = x.fi;
		if(!vis[v]){
			dfs(v);
			cnt[u]+=cnt[v];
			cnt[u]%=2;
			if(cnt[v]%2) swap(ans[x.se].fi,ans[x.se].se); 
		}
	}
}
int main(){
	cin >> n >> m;
	for(int i=1;i<=m;i++){
		int u, v;
		cin >> u >> v;
		out[u]++;
		G[u].push_back({v,i});
		G[v].push_back({u,i});
		ans[i] = {u,v};
	}
	int tmp = 0;
	for(int i=1;i<=n;i++) out[i] %= 2, tmp+= out[i];
	if(tmp%2) return cout << -1, 0;
	dfs(1);
	for(int i=1;i<=m;i++) cout << ans[i].fi << " " << ans[i].se << endl;
}
```