#include <string.h>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

const int kMaxV = 100010;

int v, e, source = 0, dfs_clock = 0;
vector<int> g[kMaxV];
int pre[kMaxV] = {0};
int post[kMaxV] = {0};
int out_degree[kMaxV] = {0};

void DFS(int u, int father) {
  pre[u] = ++dfs_clock;
  for (int v: g[u]) {
    if (!pre[v]) {
      DFS(v, u);
    } else if (v != father && !post[v]) {
      ++out_degree[v];
      printf("%d %d\n", v, u);
    }
  }
  if (father <= 0) {
    return;
  }
  post[u] = ++dfs_clock;
  if (out_degree[u] % 2 == 0) {
    ++out_degree[father];
    printf("%d %d\n", father, u);
  } else {
    ++out_degree[u];
    printf("%d %d\n", u, father);
  }
}

int main() {
  scanf("%d %d", &v, &e);
  for (int i = 0; i < e; ++i) {
    int s, t;
    scanf("%d %d", &s, &t);
    g[s].push_back(t);
    g[t].push_back(s);
  }
  if (e % 2 != 0) {
    printf("%d\n", -1);
  } else {
    DFS(1, -1);
  }

  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <string.h>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

const int kMaxV = 100010;

int v, e, source = 0, dfs_clock = 0;
vector<int> g[kMaxV];
int pre[kMaxV] = {0};
int post[kMaxV] = {0};
int out_degree[kMaxV] = {0};

void DFS(int u, int father) {
  pre[u] = ++dfs_clock;
  for (int v: g[u]) {
    if (!pre[v]) {
      DFS(v, u);
    } else if (v != father && !post[v]) {
      ++out_degree[v];
      printf("%d %d\n", v, u);
    }
  }
  if (father <= 0) {
    return;
  }
  post[u] = ++dfs_clock;
  if (out_degree[u] % 2 == 0) {
    ++out_degree[father];
    printf("%d %d\n", father, u);
  } else {
    ++out_degree[u];
    printf("%d %d\n", u, father);
  }
}

int main() {
  scanf("%d %d", &v, &e);
  for (int i = 0; i < e; ++i) {
    int s, t;
    scanf("%d %d", &s, &t);
    g[s].push_back(t);
    g[t].push_back(s);
  }
  if (e % 2 != 0) {
    printf("%d\n", -1);
  } else {
    DFS(1, -1);
  }

  return 0;
}

```