#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int N = 1e5 + 5;
vector<int> g[N];
int p[N], col[N];

void dfs(int node, int anc) {
    col[node] = 1;
    for (int to : g[node]) {
        if (!col[to]) {
            dfs(to, node);

            if (p[to]) {
                cout << to + 1 << " " << node + 1 << "\n";
            }
            else {
                p[node] ^= 1;
                cout << node + 1 << " " << to + 1 << "\n";
            }
        }
        else if (to != anc && col[to] == 1) {
            p[node] ^= 1;
            cout << node + 1 << " " << to + 1 << "\n";
        }
    }
    col[node] = 2;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m;
    cin >> n >> m;

    for (int i = 0; i < m; i++) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }

    if (m % 2) {
        cout << "-1\n";
    }
    else {
        dfs(0, -1);
    }

    return 0;
}
### Prompt

Please formulate 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;
 
#define ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int N = 1e5 + 5;
vector<int> g[N];
int p[N], col[N];

void dfs(int node, int anc) {
    col[node] = 1;
    for (int to : g[node]) {
        if (!col[to]) {
            dfs(to, node);

            if (p[to]) {
                cout << to + 1 << " " << node + 1 << "\n";
            }
            else {
                p[node] ^= 1;
                cout << node + 1 << " " << to + 1 << "\n";
            }
        }
        else if (to != anc && col[to] == 1) {
            p[node] ^= 1;
            cout << node + 1 << " " << to + 1 << "\n";
        }
    }
    col[node] = 2;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m;
    cin >> n >> m;

    for (int i = 0; i < m; i++) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }

    if (m % 2) {
        cout << "-1\n";
    }
    else {
        dfs(0, -1);
    }

    return 0;
}
```