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

Solution (in CPP):

Solution:
// written by newbiechd
#include <algorithm>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>

// DELETE the debugging information!
#define debug(x) std::cerr << #x << " = " << (x) << std::endl

int read() {
  int f = 0;
  char ch;
  while (!isdigit(ch = getchar()));
  do f = 10 * f + (ch ^ '0');
  while (isdigit(ch = getchar()));
  return f;
}

const int N_MAX = 100003;
std::vector<int> g[N_MAX];
int dep[N_MAX];
bool deg[N_MAX];
void dfs(int x, int f) {
  dep[x] = dep[f] + 1;
  for (int y : g[x])
    if (y != f) {
      if (dep[y]) {
        if (dep[x] > dep[y])
          printf("%d %d\n", x, y), deg[x] ^= 1;
      }
      else
        dfs(y, x);
    }
  if (x > 1) {
    if (deg[x])
      printf("%d %d\n", x, f), deg[x] = 0;
    else
      printf("%d %d\n", f, x), deg[f] ^= 1;
  }
}

int main() {
  int n = read(), m = read();
  for (int i = 1; i <= m; ++i) {
    int x = read(), y = read();
    g[x].push_back(y), g[y].push_back(x);
  }
  if (m & 1)
    printf("-1\n");
  else
    dfs(1, 0);
  return 0;
}