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

typedef long long ll;
#define forn(i,n) for (int i = 0; i < int(n); i++)

const int N = 1e5 + 10;

int n;
bool odd[N], rev[N], vis[N];
vector<int> adj[N];
vector<pair<int,int>> edge;

void dfs(int u) {
    vis[u] = true;
    for (auto idx : adj[u]) {
        auto &e  = edge[idx];
        int v = e.first + e.second - u;
        if (!vis[v]) {
            dfs(v);
            if (odd[v]) {
                odd[v] = !odd[v];
                odd[u] = !odd[u];
                rev[idx] = true;
            }
        }
    }
}

int main() {
    int m;
    cin >> n >> m;
    if (m%2) {
        cout << -1;
        return 0;
    }
    forn(i,m) {
        int u,v; cin >> u >> v;
        u--; v--;
        edge.emplace_back(u,v);
        adj[u].push_back(i);
        adj[v].push_back(i);
        odd[u] = !odd[u];
    }
    dfs(0);
    forn(i,m) {
        int u,v; tie(u,v) = edge[i];
        if (rev[i]) swap(u,v);
        cout << u+1 << ' ' << v+1 << '\n';
    }

    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;
#define forn(i,n) for (int i = 0; i < int(n); i++)

const int N = 1e5 + 10;

int n;
bool odd[N], rev[N], vis[N];
vector<int> adj[N];
vector<pair<int,int>> edge;

void dfs(int u) {
    vis[u] = true;
    for (auto idx : adj[u]) {
        auto &e  = edge[idx];
        int v = e.first + e.second - u;
        if (!vis[v]) {
            dfs(v);
            if (odd[v]) {
                odd[v] = !odd[v];
                odd[u] = !odd[u];
                rev[idx] = true;
            }
        }
    }
}

int main() {
    int m;
    cin >> n >> m;
    if (m%2) {
        cout << -1;
        return 0;
    }
    forn(i,m) {
        int u,v; cin >> u >> v;
        u--; v--;
        edge.emplace_back(u,v);
        adj[u].push_back(i);
        adj[v].push_back(i);
        odd[u] = !odd[u];
    }
    dfs(0);
    forn(i,m) {
        int u,v; tie(u,v) = edge[i];
        if (rev[i]) swap(u,v);
        cout << u+1 << ' ' << v+1 << '\n';
    }

    return 0;
}

```