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

typedef long long               ll;
typedef unsigned long long      ull;

const int N = 100 * 1000 + 20;

vector <ll> adj[N];
bool vis[N];
ll d[N], outD[N];

void dfs(ll v) {
    vis[v] = true;
    for (auto u: adj[v]) {
        if (!vis[u]) {
            d[u] = d[v] + 1;
            dfs(u);
        }
        if (d[u] > d[v]) {
            if (!(outD[u] & 1)) {
                outD[v]++;
                cout << v << " " << u << endl;
            }
            else {
                outD[u]++;
                cout << u << ' '  << v << endl;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    ll n, m;
    cin >> n >> m;
    if (m & 1)
        return cout << -1 << endl, 0;
    for (int i = 0, u, v; i < m; i++) {
        cin >> v >> u;
        adj[v].push_back(u);
        adj[u].push_back(v);
    }
    dfs(1);
    return 0;
}

### 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>
using namespace std;

typedef long long               ll;
typedef unsigned long long      ull;

const int N = 100 * 1000 + 20;

vector <ll> adj[N];
bool vis[N];
ll d[N], outD[N];

void dfs(ll v) {
    vis[v] = true;
    for (auto u: adj[v]) {
        if (!vis[u]) {
            d[u] = d[v] + 1;
            dfs(u);
        }
        if (d[u] > d[v]) {
            if (!(outD[u] & 1)) {
                outD[v]++;
                cout << v << " " << u << endl;
            }
            else {
                outD[u]++;
                cout << u << ' '  << v << endl;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    ll n, m;
    cin >> n >> m;
    if (m & 1)
        return cout << -1 << endl, 0;
    for (int i = 0, u, v; i < m; i++) {
        cin >> v >> u;
        adj[v].push_back(u);
        adj[u].push_back(v);
    }
    dfs(1);
    return 0;
}

```