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

int n, m;
const int maxn = 1e5 + 10;
vector<int> g[maxn];
vector<pair<int, int>> e;

bool v[maxn], ve[maxn];
int rev[maxn];

int dfs(int x) {
  v[x] = true;
  int r = 0;
  for (int t : g[x]) {
    if (ve[t]) {
      continue;
    }
    ve[t] = true;
    r ^= 1;
    auto edge = e[t];
    if (x == edge.second) rev[t] = 1;
    int y = x == edge.first ? edge.second : edge.first;
    if (v[y]) {
      continue;
    }
    if (dfs(y)) {
      r ^= 1;
      rev[t] ^= 1;
    }
  }
  return r;
}


int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, u, v; i < m ; ++i) {
    scanf("%d%d", &u, &v);
    g[u].push_back(e.size());
    g[v].push_back(e.size());
    e.emplace_back(u, v);
  }
  if (dfs(1)) {
    puts("-1");
    return 0;
  }
  for (int i = 0; i < m; ++i) {
    if (rev[i])
      swap(e[i].first, e[i].second);
    printf("%d %d\n", e[i].first, e[i].second);
  }

  return 0;
}