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 <algorithm>
#include <cstdio>
#include <utility>
#include <vector>

using namespace std;

const int N = 100000 + 5;

bool vis[N];
int c[N], ex[N], ey[N];
vector<pair<int, int>> v[N];

void dfs(int x)
{
    vis[x] = true;

    for (pair<int, int> pr : v[x])
    {
        if (vis[pr.first])
        {
            continue;
        }

        dfs(pr.first);

        if (c[pr.first] == 1)
        {
            c[pr.first] ^= 1;
            c[x] ^= 1;
            swap(ex[pr.second], ey[pr.second]);
        }
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);

    if (m % 2 != 0)
    {
        puts("-1");
        return 0;
    }

    for (int i = 0; i < m; i++)
    {
        scanf("%d%d", &ex[i], &ey[i]);
        v[ex[i]].push_back(make_pair(ey[i], i));
        v[ey[i]].push_back(make_pair(ex[i], i));
        c[ex[i]] ^= 1;
    }

    dfs(1);

    for (int i = 0; i < m; i++)
    {
        printf("%d %d\n", ex[i], ey[i]);
    }

    return 0;
}