#include<bits/stdc++.h>

using namespace std;

int N, M, xo[100009], ap[100009], t[100009];
vector < int > v[100009], ord;

void dfs (int nod, int tata)
{
    ap[nod] = 2;
    for (auto it : v[nod])
        if (ap[it] == 0)
            t[it] = nod, dfs (it, nod);
        else
        if (it != tata && ap[it] == 2)
        {
            printf ("%d %d\n", it, nod);
            xo[it] ^= 1;
        }
    ap[nod] = 1;
    ord.push_back (nod);
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d", &N, &M);
if (M & 1) printf ("-1\n");
else
{
    for (int i=1; i<=M; i++)
    {
        int x, y;
        scanf ("%d %d", &x, &y);
        v[x].push_back (y);
        v[y].push_back (x);
    }
    dfs (1, -1);
    for (auto it : ord)
        if (it != 1)
        {
            if (xo[it]) printf ("%d %d\n", it, t[it]);
            else printf ("%d %d\n", t[it], it), xo[t[it]] ^= 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;

int N, M, xo[100009], ap[100009], t[100009];
vector < int > v[100009], ord;

void dfs (int nod, int tata)
{
    ap[nod] = 2;
    for (auto it : v[nod])
        if (ap[it] == 0)
            t[it] = nod, dfs (it, nod);
        else
        if (it != tata && ap[it] == 2)
        {
            printf ("%d %d\n", it, nod);
            xo[it] ^= 1;
        }
    ap[nod] = 1;
    ord.push_back (nod);
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d %d", &N, &M);
if (M & 1) printf ("-1\n");
else
{
    for (int i=1; i<=M; i++)
    {
        int x, y;
        scanf ("%d %d", &x, &y);
        v[x].push_back (y);
        v[y].push_back (x);
    }
    dfs (1, -1);
    for (auto it : ord)
        if (it != 1)
        {
            if (xo[it]) printf ("%d %d\n", it, t[it]);
            else printf ("%d %d\n", t[it], it), xo[t[it]] ^= 1;
        }
}
return 0;
}

```