#include <bits/stdc++.h>

const int N = 1e5 + 233;

int n, m, tot, vis[N], out[N];

struct Edge { int x, y; } E[N];
struct Node { int y, id, t; };
std::vector<Node> G[N];

void dfs1(int x, int fa) {
  vis[x] = 1;
  for (int i = 0; i < G[x].size(); ++i) {
    int y = G[x][i].y;
    if (y == fa) G[x][i].t = 1;
    else if (!vis[y]) {
      G[x][i].t = 1; dfs1(y, x);
    } else if (!E[G[x][i].id].x) {
      E[G[x][i].id] = { x, y };
      ++out[x];
    }
  }
}

void dfs2(int x, int fa) {
  for (auto w : G[x]) {
    if (!w.t || w.y == fa) continue;
    int y = w.y; dfs2(y, x);
    if (out[y] & 1)
      E[w.id] = { y, x };
    else {
      ++out[x];
      E[w.id] = { x, y };
    }
  }
}

int main() {
  scanf("%d %d", &n, &m);
  if (m & 1) { puts("-1"); return 0; }
  for (int i = 1; i <= m; ++i) {
    int x, y; scanf("%d %d", &x, &y);
    G[x].push_back({ y, i, 0 });
    G[y].push_back({ x, i, 0 });
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= m; ++i)
    printf("%d %d\n", E[i].x, E[i].y);
  return 0;
}

### Prompt

In CPP, your task is to solve 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>

const int N = 1e5 + 233;

int n, m, tot, vis[N], out[N];

struct Edge { int x, y; } E[N];
struct Node { int y, id, t; };
std::vector<Node> G[N];

void dfs1(int x, int fa) {
  vis[x] = 1;
  for (int i = 0; i < G[x].size(); ++i) {
    int y = G[x][i].y;
    if (y == fa) G[x][i].t = 1;
    else if (!vis[y]) {
      G[x][i].t = 1; dfs1(y, x);
    } else if (!E[G[x][i].id].x) {
      E[G[x][i].id] = { x, y };
      ++out[x];
    }
  }
}

void dfs2(int x, int fa) {
  for (auto w : G[x]) {
    if (!w.t || w.y == fa) continue;
    int y = w.y; dfs2(y, x);
    if (out[y] & 1)
      E[w.id] = { y, x };
    else {
      ++out[x];
      E[w.id] = { x, y };
    }
  }
}

int main() {
  scanf("%d %d", &n, &m);
  if (m & 1) { puts("-1"); return 0; }
  for (int i = 1; i <= m; ++i) {
    int x, y; scanf("%d %d", &x, &y);
    G[x].push_back({ y, i, 0 });
    G[y].push_back({ x, i, 0 });
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= m; ++i)
    printf("%d %d\n", E[i].x, E[i].y);
  return 0;
}

```