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

const int N = 1e5 + 100;

vector<int> g[N];
int depth[N];
int vis[N];

int dfs(int u, int p, int d = 0) {
    int cnt = 0;
    depth[u] = d;
    vis[u] = 1;
    for(int v : g[u]) {
        if(v == p) continue;
        if(vis[v] && depth[v] < depth[u]) {
            // back edge
            cout << u << " " << v << "\n";
            cnt++;
        }
        if(vis[v] == 0) {
            if(dfs(v, u, d+1)) {
                cout << u << " " << v << "\n";
                cnt++;
            }
        }
    }
    if(cnt&1) {
        cout << u << " " << p << endl;
        return 0;
    }
    return 1;
}

int main() {
    // freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, m;
    cin >> n >> m;
    if(m&1) {
        cout << -1 << endl;
        return 0;
    }
    for(int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    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 N = 1e5 + 100;

vector<int> g[N];
int depth[N];
int vis[N];

int dfs(int u, int p, int d = 0) {
    int cnt = 0;
    depth[u] = d;
    vis[u] = 1;
    for(int v : g[u]) {
        if(v == p) continue;
        if(vis[v] && depth[v] < depth[u]) {
            // back edge
            cout << u << " " << v << "\n";
            cnt++;
        }
        if(vis[v] == 0) {
            if(dfs(v, u, d+1)) {
                cout << u << " " << v << "\n";
                cnt++;
            }
        }
    }
    if(cnt&1) {
        cout << u << " " << p << endl;
        return 0;
    }
    return 1;
}

int main() {
    // freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, m;
    cin >> n >> m;
    if(m&1) {
        cout << -1 << endl;
        return 0;
    }
    for(int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1, -1);

    return 0;
}


```