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

struct Edge {
    int u, v;
    bool dir;

    int other(int x)    {return x^u^v;}
};

const int N = 1e5+7;
vector<int> adj[N];
vector<Edge> edges;
bool deg[N];
bool vis[N];

void dfs(int u) {
    vis[u] = 1;
    for (int id: adj[u]) {
        int v = edges[id].other(u);
        if (vis[v])     continue;
        dfs(v);
        if (deg[v]) {
            edges[id].dir = 1;
            deg[v] = !deg[v];
            deg[u] = !deg[u];
        }
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, m;
    cin>>n>>m;

    if (m%2) {
        cout<<-1<<endl;
        return 0;
    }

    for (int i=0; i<m; i++) {
        int u, v;
        cin>>u>>v;
        edges.push_back({u, v, 0});
        adj[u].push_back(edges.size()-1);
        adj[v].push_back(edges.size()-1);
        deg[u] ^= 1;
    }

    dfs(1);
    assert(deg[1] == 0);

    for (Edge e: edges)
        if (e.dir)  cout<<e.v<<" "<<e.u<<"\n";
        else        cout<<e.u<<" "<<e.v<<"\n";
}

### 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;

struct Edge {
    int u, v;
    bool dir;

    int other(int x)    {return x^u^v;}
};

const int N = 1e5+7;
vector<int> adj[N];
vector<Edge> edges;
bool deg[N];
bool vis[N];

void dfs(int u) {
    vis[u] = 1;
    for (int id: adj[u]) {
        int v = edges[id].other(u);
        if (vis[v])     continue;
        dfs(v);
        if (deg[v]) {
            edges[id].dir = 1;
            deg[v] = !deg[v];
            deg[u] = !deg[u];
        }
    }
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, m;
    cin>>n>>m;

    if (m%2) {
        cout<<-1<<endl;
        return 0;
    }

    for (int i=0; i<m; i++) {
        int u, v;
        cin>>u>>v;
        edges.push_back({u, v, 0});
        adj[u].push_back(edges.size()-1);
        adj[v].push_back(edges.size()-1);
        deg[u] ^= 1;
    }

    dfs(1);
    assert(deg[1] == 0);

    for (Edge e: edges)
        if (e.dir)  cout<<e.v<<" "<<e.u<<"\n";
        else        cout<<e.u<<" "<<e.v<<"\n";
}

```