#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, b) for (int i = (a); i < (b); i++)

int n, m;
vector<vector<int>> g(101000);
vector<int> deg(101000), depth(101000, -1);
vector<pair<int, int>> ans;

void dfs(int v, int p) {
    for (auto u : g[v]) {
        if (depth[u] == -1) {
            depth[u] = depth[v] + 1;
            dfs(u, v);
        } else if (depth[u] > depth[v]) {
            ans.emplace_back(v, u);
            deg[v]++;
        }
    }
    if (p != -1) {
        if (deg[v] % 2 == 1) {
            ans.emplace_back(v, p);
            deg[v]++;
        } else {
            ans.emplace_back(p, v);
            deg[p]++;
        }
    }
    return;
}

int main() {
    cin >> n >> m;
    if (m % 2) {
        cout << -1 << endl;
        return 0;
    }
    rep(i, 0, m) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        g[x].emplace_back(y);
        g[y].emplace_back(x);
    }
    depth[0] = 0;
    dfs(0, -1);

    for (auto p : ans) {
        cout << p.first + 1 << ' ' << p.second + 1 << endl;
    }
}
### Prompt

Your task is to create a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, a, b) for (int i = (a); i < (b); i++)

int n, m;
vector<vector<int>> g(101000);
vector<int> deg(101000), depth(101000, -1);
vector<pair<int, int>> ans;

void dfs(int v, int p) {
    for (auto u : g[v]) {
        if (depth[u] == -1) {
            depth[u] = depth[v] + 1;
            dfs(u, v);
        } else if (depth[u] > depth[v]) {
            ans.emplace_back(v, u);
            deg[v]++;
        }
    }
    if (p != -1) {
        if (deg[v] % 2 == 1) {
            ans.emplace_back(v, p);
            deg[v]++;
        } else {
            ans.emplace_back(p, v);
            deg[p]++;
        }
    }
    return;
}

int main() {
    cin >> n >> m;
    if (m % 2) {
        cout << -1 << endl;
        return 0;
    }
    rep(i, 0, m) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        g[x].emplace_back(y);
        g[y].emplace_back(x);
    }
    depth[0] = 0;
    dfs(0, -1);

    for (auto p : ans) {
        cout << p.first + 1 << ' ' << p.second + 1 << endl;
    }
}
```