#include <bits/stdc++.h>
#define f first
#define s second
using namespace std;
using ll = long long;
using ii = pair<int, int>;

int c[100001];
vector<int> G[100001];
bitset<100001> vis;
set<ii> used;
vector<ii> ans;

inline void mark(int u, int v) { ans.emplace_back(u, v), ++c[u], used.emplace(u, v), used.emplace(v, u); }

void dfs(int u, int p) {
    vis[u] = 1;
    for (int v : G[u]) if (v != p && !used.count(ii(u, v))) {
        if (!vis[v]) {
            dfs(v, u);
            if (c[v]&1) mark(v, u);
            else mark(u, v);
        }
        else mark(u, v);
    }
}

int main() {
    if (fopen("in", "r")) freopen("in", "r", stdin), freopen("out", "w", stdout);
    ios_base::sync_with_stdio(0), cin.tie(0);

    int N, M; cin >> N >> M;
    if (M&1) {
        cout << -1;
        return 0;
    }
    while (M--) {
        int A, B; cin >> A >> B;
        G[A].push_back(B), G[B].push_back(A);
    }

    dfs(1, 0);
    
    for (ii p : ans) cout << p.f << ' ' << p.s << '\n';
    return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
#define f first
#define s second
using namespace std;
using ll = long long;
using ii = pair<int, int>;

int c[100001];
vector<int> G[100001];
bitset<100001> vis;
set<ii> used;
vector<ii> ans;

inline void mark(int u, int v) { ans.emplace_back(u, v), ++c[u], used.emplace(u, v), used.emplace(v, u); }

void dfs(int u, int p) {
    vis[u] = 1;
    for (int v : G[u]) if (v != p && !used.count(ii(u, v))) {
        if (!vis[v]) {
            dfs(v, u);
            if (c[v]&1) mark(v, u);
            else mark(u, v);
        }
        else mark(u, v);
    }
}

int main() {
    if (fopen("in", "r")) freopen("in", "r", stdin), freopen("out", "w", stdout);
    ios_base::sync_with_stdio(0), cin.tie(0);

    int N, M; cin >> N >> M;
    if (M&1) {
        cout << -1;
        return 0;
    }
    while (M--) {
        int A, B; cin >> A >> B;
        G[A].push_back(B), G[B].push_back(A);
    }

    dfs(1, 0);
    
    for (ii p : ans) cout << p.f << ' ' << p.s << '\n';
    return 0;
}
```