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 = 1e5 + 10;
int n, m;
int v[N][2];
int vise[N], visv[N];
vector<int> adj[N];

bool DFS(int u, int pa) {
    visv[u] = 1;
    int n1 = 0;
    for(auto i : adj[u]) if (!vise[i]) {
        vise[i] = 1;
        int v = ::v[i][0] + ::v[i][1] - u;
        if (visv[v]) {
            cout << u << " " << v << '\n';
            n1++;
        } else {
            if (!DFS(v, u)) {
                cout << u << " " << v << '\n';
                n1++;
            }else cout << v << " " << u << '\n';
        }
    }
    return n1 % 2;
}
int main() {
    cin >> n >> m;
    if (m % 2) {
        cout << "-1";
        return 0;
    }
    for(int i = 1; i <= m; i++) {
        cin >> v[i][0] >> v[i][1];
        adj[v[i][0]].push_back(i);
        adj[v[i][1]].push_back(i);
    }
    assert(!DFS(1, -1));
}