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

const int MAX_N = (int)1e5 + 5;
int depth[MAX_N], deg_out[MAX_N];
vector<int> G[MAX_N];

void dfs(int s, int p){
    for(auto u : G[s]){
        if(u==p) continue;
        if(depth[u]==-1){
            depth[u] = depth[s] + 1;
            dfs(u, s);
        } else if(depth[u]<depth[s]){
            printf("%d %d\n", u, s);
            deg_out[u]++;
        }
    }
    if(p==-1) return;
    if(deg_out[s]%2){
        printf("%d %d\n", s, p);
        deg_out[s]++;
    } else {
        printf("%d %d\n", p, s);
        deg_out[p]++;
    }
}

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int N, M;
    cin>>N>>M;
    if(M%2){cout<<"-1"<<endl; return 0;}
    for(int i=0;i<M;i++){
        int a, b;
        cin>>a>>b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    fill(depth, depth+N+1, -1);
    depth[1] = 0;
    dfs(1, -1);
    return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 MAX_N = (int)1e5 + 5;
int depth[MAX_N], deg_out[MAX_N];
vector<int> G[MAX_N];

void dfs(int s, int p){
    for(auto u : G[s]){
        if(u==p) continue;
        if(depth[u]==-1){
            depth[u] = depth[s] + 1;
            dfs(u, s);
        } else if(depth[u]<depth[s]){
            printf("%d %d\n", u, s);
            deg_out[u]++;
        }
    }
    if(p==-1) return;
    if(deg_out[s]%2){
        printf("%d %d\n", s, p);
        deg_out[s]++;
    } else {
        printf("%d %d\n", p, s);
        deg_out[p]++;
    }
}

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int N, M;
    cin>>N>>M;
    if(M%2){cout<<"-1"<<endl; return 0;}
    for(int i=0;i<M;i++){
        int a, b;
        cin>>a>>b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    fill(depth, depth+N+1, -1);
    depth[1] = 0;
    dfs(1, -1);
    return 0;
}
```