#include <bits/stdc++.h>
#define NMAX (int)(1e5+4)
#define pb push_back
#define ft first
#define sd second
using namespace std;
typedef pair <int, int> pairINT;
typedef long long ll;

int n,m;
bool used[NMAX],odd[NMAX];
vector <int> g[NMAX];
vector <pairINT> sol;

void dfs(int);

int main(){
    int x,y;
    cin>>n>>m;
    for(int i=0;i<m;++i){
        cin>>x>>y;
        g[x].pb(y);
        g[y].pb(x);
    }
    if(m&1){
        cout<<-1;
        return 0;
    }
    dfs(1);
    for(auto it:sol)
        cout<<it.ft<<' '<<it.sd<<'\n';
    return 0;
}
void dfs(int node){
    vector <int> adj;
    used[node]=1;
    for(auto it:g[node])
        if(used[it])
            adj.push_back(it);
    for(auto it:g[node])
        if(!used[it])
            dfs(it);

    if(odd[node]){
        sol.pb({node, adj.back()});
        adj.pop_back();
    }
    for(auto it:adj)
        odd[it]^=1,
        sol.pb({it, node});
}

### Prompt

Your task is to create 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 NMAX (int)(1e5+4)
#define pb push_back
#define ft first
#define sd second
using namespace std;
typedef pair <int, int> pairINT;
typedef long long ll;

int n,m;
bool used[NMAX],odd[NMAX];
vector <int> g[NMAX];
vector <pairINT> sol;

void dfs(int);

int main(){
    int x,y;
    cin>>n>>m;
    for(int i=0;i<m;++i){
        cin>>x>>y;
        g[x].pb(y);
        g[y].pb(x);
    }
    if(m&1){
        cout<<-1;
        return 0;
    }
    dfs(1);
    for(auto it:sol)
        cout<<it.ft<<' '<<it.sd<<'\n';
    return 0;
}
void dfs(int node){
    vector <int> adj;
    used[node]=1;
    for(auto it:g[node])
        if(used[it])
            adj.push_back(it);
    for(auto it:g[node])
        if(!used[it])
            dfs(it);

    if(odd[node]){
        sol.pb({node, adj.back()});
        adj.pop_back();
    }
    for(auto it:adj)
        odd[it]^=1,
        sol.pb({it, node});
}

```