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

const int maxn = 100 * 1000 + 20;
int n, m, dis[maxn];
vector <int> adj[maxn], adjexit[maxn];
bool vis[maxn];

void input() {
    cin >> n >> m;
    for (int i = 0, a, b; i < m; i++) {
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
}

void dfs(int v) {
    vis[v] = 1;
    for (int u : adj[v]) {
        if (!vis[u]) {
            dis[u] = dis[v] + 1;
            dfs(u);
            adjexit[u].size() % 2? adjexit[u].push_back(v): adjexit[v].push_back(u);
        }
        else if (dis[v] < dis[u])
            adjexit[v].push_back(u);
    }
}
void output() {
    for (int i = 1; i <= n; i++)
        for (int j : adjexit[i])
            cout << i << ' ' << j << endl;
}
int main() {    
    ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);
    input();
    if (m % 2)
        return cout << -1 << endl, 0;
    dfs(1);
    output();
    return 0;    
}


### Prompt

Create a solution in cpp for 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;

const int maxn = 100 * 1000 + 20;
int n, m, dis[maxn];
vector <int> adj[maxn], adjexit[maxn];
bool vis[maxn];

void input() {
    cin >> n >> m;
    for (int i = 0, a, b; i < m; i++) {
        cin >> a >> b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
}

void dfs(int v) {
    vis[v] = 1;
    for (int u : adj[v]) {
        if (!vis[u]) {
            dis[u] = dis[v] + 1;
            dfs(u);
            adjexit[u].size() % 2? adjexit[u].push_back(v): adjexit[v].push_back(u);
        }
        else if (dis[v] < dis[u])
            adjexit[v].push_back(u);
    }
}
void output() {
    for (int i = 1; i <= n; i++)
        for (int j : adjexit[i])
            cout << i << ' ' << j << endl;
}
int main() {    
    ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);
    input();
    if (m % 2)
        return cout << -1 << endl, 0;
    dfs(1);
    output();
    return 0;    
}


```