Problem:
Score : 700 points

Problem Statement

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.

Notes

An undirected graph is said to be simple when it contains no self-loops or multiple edges.

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.

Sample Input 1

4 4
1 2
2 3
3 4
4 1

Sample Output 1

1 2
1 4
3 2
3 4

After this assignment of directions, Vertex 1 and 3 will each have two outgoing edges, and Vertex 2 and 4 will each have zero outgoing edges.

Sample Input 2

5 5
1 2
2 3
3 4
2 5
4 5

Sample Output 2

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using LL = long long; using PII = pair<LL, LL>; using VI = vector<LL>; using VVI = vector<VI>;
using VB = vector<bool>; using VS = vector<string>; using VP = vector<PII>;
#define VV(T)        vector<vector<T>>
#define PB           push_back
#define MP           make_pair
#define SZ(a)        LL((a).size())
#define EACH(x, c)   for (auto x : (c))
#define ALL(c)       (c).begin(), (c).end()
#define REVERSE(c)   reverse(ALL(c))
#define SORT(c)      stable_sort(ALL(c))
#define RSORT(c)     stable_sort((c).rbegin(), (c).rend())
#define FOR(i, a, b) for (LL i = (a); i < (b); ++i)
#define REP(i, n)    FOR(i, 0, n)
#define $(x)         {cout << #x << " = " << (x) << endl;}

void solve(LL N, LL M, vector<LL> A, vector<LL> B){
    if (M % 2 == 1) {
        cout << "-1" << endl;
        return;
    }
    vector<unordered_set<LL>> G(N + 1);
    REP(i, M) {
        G[A[i]].insert(B[i]);
        G[B[i]].insert(A[i]);
    }

    /*FOR(i, 1, N + 1) {
        $(i);
        //cout << *G[i].begin() << endl;
        auto hoge = *G[i].begin();
        $(hoge);
        G[i].erase(hoge);
        EACH(x, G[i]) {
            cout << x << " ";
        }
        cout << endl;
        }*/

    // determine from less out-degree node
    priority_queue<PII, VP, greater<PII>> q;   // PII = (out-degree, node)
    FOR(i, 1, N + 1) {
        q.push(MP(SZ(G[i]), i));
    }
    VB used(N + 1);
    VI out_degs(N + 1);   // out-degree of each node so far
    VP edges;
    while (!q.empty()) {
        auto v = q.top(); q.pop();
        //$(v.first);
        //$(v.second);
        if (!used[v.second]) {
            used[v.second] = true;
            // add outgoing edges as much as it has even num of out-edges
            LL n_edges_out = (out_degs[v.second] % 2 == 0) ? ((v.first / 2) * 2) : ((v.first % 2 == 0) ? v.first - 1 : v.first);
            //$(n_edges_out);
            REP(i, n_edges_out) {
                //cout << "outgoing" << endl;
                auto itr = G[v.second].begin();
                advance(itr, i);
                auto w = *itr;
                //$(w);
                edges.PB(MP(v.second, w));
                G[w].erase(v.second);
                q.push(MP(SZ(G[w]), w));
            }
            // and then add incoming edges for remaining (single) edge if remain
            if (v.first - n_edges_out == 1) {
                //cout << "incoming" << endl;
                /*EACH(x, G[v.second]) {
                    $(x);
                    }*/
                auto itr = G[v.second].begin();
                advance(itr, n_edges_out);
                auto w = *itr;
                //$(w);
                edges.PB(MP(w, v.second));
                //cout << "1\n";
                G[w].erase(v.second);
                //cout << "2\n";
                q.push(MP(SZ(G[w]), w));
                //cout << "3\n";
                out_degs[w]++;
                //cout << "4\n";
            }
        }
    }
    EACH(edge, edges) {
        cout << edge.first << " " << edge.second << endl;
    }
}

int main(){
    LL N;
    scanf("%lld",&N);
    LL M;
    scanf("%lld",&M);
    vector<LL> A(M);
    vector<LL> B(M);
    for(int i = 0 ; i < M ; i++){
        scanf("%lld",&A[i]);
        scanf("%lld",&B[i]);
    }
    solve(N, M, move(A), move(B));
    return 0;
}