#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

const int N = 1e5 + 5;
vector<int> adj[N];
int n, m, l[N], r[N], deg[N], vis[N], tree[N], par[N], d[N];

void dfs(int u) {
  vis[u] = 1;
  for (int i : adj[u]) {
    int v = u ^ l[i] ^ r[i];
    if (!vis[v]) {
      par[v] = i;
      tree[i] = 1;
      d[v] = d[u] + 1;
      dfs(v);
    }
  }
}

int main() {
  scanf("%d %d", &n, &m);
  if(m & 1) {
    puts("-1");
    exit(0); 
  }

  for (int i = 1; i <= m; ++i) {
    scanf("%d %d", &l[i], &r[i]);
    adj[l[i]].push_back(i);
    adj[r[i]].push_back(i);
  }

  dfs(1);

  for (int i = 1; i <= m; ++i) {
    if (!tree[i]) deg[l[i]] ^= 1;
  }

  vector<int> id(n - 1);
  iota(id.begin(), id.end(), 2);
  sort(id.begin(), id.end(), [](int i, int j) { return d[i] > d[j]; });

  for (int u : id) {
    int e = par[u];
    if(deg[u] ^ (l[e] == u)) swap(l[e], r[e]); 
    deg[l[e]] ^= 1;
  }

  for (int i = 1; i <= m; ++i) {
    printf("%d %d\n", l[i], r[i]);
  }
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;

typedef long long ll;
typedef pair<int, int> ii;

const int N = 1e5 + 5;
vector<int> adj[N];
int n, m, l[N], r[N], deg[N], vis[N], tree[N], par[N], d[N];

void dfs(int u) {
  vis[u] = 1;
  for (int i : adj[u]) {
    int v = u ^ l[i] ^ r[i];
    if (!vis[v]) {
      par[v] = i;
      tree[i] = 1;
      d[v] = d[u] + 1;
      dfs(v);
    }
  }
}

int main() {
  scanf("%d %d", &n, &m);
  if(m & 1) {
    puts("-1");
    exit(0); 
  }

  for (int i = 1; i <= m; ++i) {
    scanf("%d %d", &l[i], &r[i]);
    adj[l[i]].push_back(i);
    adj[r[i]].push_back(i);
  }

  dfs(1);

  for (int i = 1; i <= m; ++i) {
    if (!tree[i]) deg[l[i]] ^= 1;
  }

  vector<int> id(n - 1);
  iota(id.begin(), id.end(), 2);
  sort(id.begin(), id.end(), [](int i, int j) { return d[i] > d[j]; });

  for (int u : id) {
    int e = par[u];
    if(deg[u] ^ (l[e] == u)) swap(l[e], r[e]); 
    deg[l[e]] ^= 1;
  }

  for (int i = 1; i <= m; ++i) {
    printf("%d %d\n", l[i], r[i]);
  }
}

```