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

const int NMAX = 1e5 + 5;

int n, m;
int depth[NMAX];
vector <int> G[NMAX];
int curr[NMAX];

void dfs(int nod, int parent) {
    for (auto v: G[nod]) {
        if (v == parent)
            continue;

        if (depth[v]) {
            // muchie de intoarcere
            if (depth[v] < depth[nod]) {
                curr[nod] ^= 1;
                cout << nod << " " << v << "\n";
            }
        }
        else {
            depth[v] = depth[nod] + 1;
            dfs(v, nod);
        }
    }

    if (curr[nod])
        cout << nod << " " << parent << "\n", curr[nod] ^= 1;
    else if (parent > 0) {
        cout << parent << " " << nod << "\n", curr[parent] ^= 1;
    }
}

int main()
{
    cin >> n >> m;

    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }

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


    depth[1] = 1;
    dfs(1, 0);

    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 NMAX = 1e5 + 5;

int n, m;
int depth[NMAX];
vector <int> G[NMAX];
int curr[NMAX];

void dfs(int nod, int parent) {
    for (auto v: G[nod]) {
        if (v == parent)
            continue;

        if (depth[v]) {
            // muchie de intoarcere
            if (depth[v] < depth[nod]) {
                curr[nod] ^= 1;
                cout << nod << " " << v << "\n";
            }
        }
        else {
            depth[v] = depth[nod] + 1;
            dfs(v, nod);
        }
    }

    if (curr[nod])
        cout << nod << " " << parent << "\n", curr[nod] ^= 1;
    else if (parent > 0) {
        cout << parent << " " << nod << "\n", curr[parent] ^= 1;
    }
}

int main()
{
    cin >> n >> m;

    for (int i = 1; i <= m; i++) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }

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


    depth[1] = 1;
    dfs(1, 0);

    return 0;
}

```