#include<bits/stdc++.h>

using namespace std;

const int   N   = 1e5 + 1;

vector<int> g[N];
vector<int> v(N);

int f[N];
int tot = 0;

int dfs(int u,int p)    {
    v[u] = ++tot;
    for(int x : g[u])   if (x != p) {
        if(!v[x])   dfs(x,u);
        if (v[x] < v[u])    {
            cout << u << " " << x << "\n";
            f[u] ^= 1;
        }
    }
    if (f[u])   cout << u << " " << p << "\n", f[u] ^= 1;
    else if (p) cout << p << " " << u << "\n", f[p] ^= 1;
}

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

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

    if (m & 1)  return  0 * puts("-1");

    while (m--) {
        int x;  cin >> x;
        int y;  cin >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(1,0);
}

### Prompt

Generate 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;

const int   N   = 1e5 + 1;

vector<int> g[N];
vector<int> v(N);

int f[N];
int tot = 0;

int dfs(int u,int p)    {
    v[u] = ++tot;
    for(int x : g[u])   if (x != p) {
        if(!v[x])   dfs(x,u);
        if (v[x] < v[u])    {
            cout << u << " " << x << "\n";
            f[u] ^= 1;
        }
    }
    if (f[u])   cout << u << " " << p << "\n", f[u] ^= 1;
    else if (p) cout << p << " " << u << "\n", f[p] ^= 1;
}

int main()  {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

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

    if (m & 1)  return  0 * puts("-1");

    while (m--) {
        int x;  cin >> x;
        int y;  cin >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(1,0);
}

```