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;

const int MAXN = 1e5 + 10;

int N, M;
vector<int> V[MAXN];
int P[MAXN], A[MAXN];

void dfs(int here, int par){
  P[here] = par;
  for(int there : V[here]) if(!P[there]) dfs(there, here);
  int s = 0;
  for(int there : V[here]) {
    if(there == par) continue;
    if(P[there] == here && A[there] == 0) s ^= 1;
    if(P[there] != here && here < there) s ^= 1;
  }
  A[here] = s;
}

void p(int x, int y){
  cout << x << ' ' << y << '\n';
}
int main(void){
  cin >> N >> M;
  if(M & 1) return cout << -1, 0;
  for(int i = 1 ; i <= M ; i++){
    int x, y;
    cin >> x >> y;
    V[x].push_back(y);   V[y].push_back(x);
  }

  dfs(1, -1);
  for(int here = 1 ; here <= N ; here++) {
    for(int there : V[here]) {
      if(P[here] == there) continue;
      if(P[there] == here) {
        if(A[there]) p(there, here);
        else p(here, there);
      }
      else {
        if(here < there) p(here, there);
      }
    }
  }

  return 0;
}