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 <functional>
#include <vector>
#include <iostream>
using namespace std;

int n, m;
vector<int> adj[100100];
pair<int, int> es[100100];
int deg[100100] = {};
bool st[100100] = {}, conn[100100] = {};

void mkst(int i) {
    conn[i] = true;
    for (int e : adj[i]) {
        int j = i^es[e].first^es[e].second;
        if (!conn[j]) { st[e] = true; mkst(j); }
    }
}

void dfs(int i, int p) {
    for (int e : adj[i]) {
        int j = i^es[e].first^es[e].second;
        if (j != p && st[e]) {
            dfs(j, i);
            if (deg[j]&1) {
                swap(es[e].first, es[e].second);
                deg[j]++; deg[i]++;
            }
        }
    }
}

int main() {
    cin >> n >> m;
    if (m % 2) { cout << "-1\n"; return 0; }
    for (int i = 0; i < m; i++) {
        int u, v; cin >> u >> v;
        u--; v--;
        es[i] = {u, v};
        adj[u].push_back(i);
        adj[v].push_back(i);
        deg[u]++;
    }
    mkst(0);
    dfs(0, -1);
    for (int i = 0; i < m; i++) cout << es[i].first+1 << ' ' << es[i].second+1 << endl;
}