#include <bits/stdc++.h>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;
typedef vector<ll> vl;

const ll MOD = 1e9 + 7;
const ll INF = 1e18;
const ll N = 1e6 + 10;

vl G[N];

map<pair<ll, ll>, ll> mp;

ll n, m, H[N], Cnt[N];

bitset<N> mark;

void DFS(ll v, ll p = 0, ll h = 0){
	mark[v] = 1;
	H[v] = h; 
	for (auto u:G[v]){
		if (!mark[u]){
			DFS(u, v, h + 1);
		}else{
			if (H[v] - 1 > H[u]) Cnt[v] ++, mp[{v, u}] = 1;
		}
	}
	if (Cnt[v] % 2 == 0 && p != 0){
		Cnt[p] ++;
		mp[{p, v}] = 1;
	}else if(Cnt[v] % 2 == 1){
		Cnt[v] ++;
		mp[{v, p}] = 1;
	}
	return;
}

int main(){
	cin >> n >> m;
	if (m % 2 == 1) return cout << -1, 0;
	for (int i = 0; i < m; i++){
		ll v, u;
		cin >> v >> u;
		G[v].pb(u), G[u].pb(v);
	}
	DFS(1);
	for (auto u:mp) cout << u.F.F << ' ' << u.F.S << '\n';




	return 0;
}

### Prompt

Please formulate 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>

#define pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;
typedef vector<ll> vl;

const ll MOD = 1e9 + 7;
const ll INF = 1e18;
const ll N = 1e6 + 10;

vl G[N];

map<pair<ll, ll>, ll> mp;

ll n, m, H[N], Cnt[N];

bitset<N> mark;

void DFS(ll v, ll p = 0, ll h = 0){
	mark[v] = 1;
	H[v] = h; 
	for (auto u:G[v]){
		if (!mark[u]){
			DFS(u, v, h + 1);
		}else{
			if (H[v] - 1 > H[u]) Cnt[v] ++, mp[{v, u}] = 1;
		}
	}
	if (Cnt[v] % 2 == 0 && p != 0){
		Cnt[p] ++;
		mp[{p, v}] = 1;
	}else if(Cnt[v] % 2 == 1){
		Cnt[v] ++;
		mp[{v, p}] = 1;
	}
	return;
}

int main(){
	cin >> n >> m;
	if (m % 2 == 1) return cout << -1, 0;
	for (int i = 0; i < m; i++){
		ll v, u;
		cin >> v >> u;
		G[v].pb(u), G[u].pb(v);
	}
	DFS(1);
	for (auto u:mp) cout << u.F.F << ' ' << u.F.S << '\n';




	return 0;
}

```