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>
#pragma comment(linker, "/STACK:16000000")
using namespace std;

typedef pair <int, int> ii;

const int Maxn = 100005;

int n, m;
vector <int> neigh[Maxn];
vector <ii> res;
int cur, tk[Maxn];
int my[Maxn];

int Solve(int v, int p = 0)
{
    tk[v] = ++cur;
    int my = 0;
    for (int i = 0; i < neigh[v].size(); i++) {
        int u = neigh[v][i];
        if (u == p) continue;
        if (!tk[u])
            if (Solve(u, v)) { my++; res.push_back(ii(v, u)); }
            else res.push_back(ii(u, v));
        else if (tk[u] < tk[v]) { my++; res.push_back(ii(v, u)); }
    }
    if (my % 2) return 0;
    return 1;
}

int main()
{
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int a, b; scanf("%d %d", &a, &b);
        neigh[a].push_back(b);
        neigh[b].push_back(a);
    }
    if (Solve(1) == 0) printf("-1\n");
    else for (int i = 0; i < res.size(); i++)
            printf("%d %d\n", res[i].first, res[i].second);
    return 0;
}