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

Solution (in CPP):

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

const int N = 100005;

int n, m, u, v, cnt = 0, in[N], bit[N];
vector<int> adj[N];

void print(int u, int v)
{
    if (u == 0 || v == 0)
        return;
    cout << u << " " << v << '\n';
    bit[u] ^= 1;
}

void DFS(int u, int p = 0)
{
    in[u] = ++cnt;
    for (int &v : adj[u])
        if (v != p)
        {
            if (in[v] == 0)
                DFS(v, u);
            else if (in[v] > in[u])
                print(u, v);
        }
    if (bit[u])
        print(u, p);
    else
        print(p, u);
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    if (m & 1)
        return cout << -1, 0;
    while (m--)
    {
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    DFS(1);
}