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>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define pb push_back

using namespace std;

vector<vector<int>> G;
vector<int> dist;

void print(int a, int b){ cout << a+1 << " " << b+1 << endl; }
void dfs(int v, int p){
    dist[v]++;
    for(auto nv : G[v]){
        if(nv == p) continue;
        if(dist[nv] >= 0){
            if(v < nv) print(v,nv), dist[v]++;
        }
        else{
            dfs(nv,v);
        }
    }
    if(p >= 0){
        if(dist[v]%2) print(v,p), dist[v]++;
        else print(p,v), dist[p]++;
    }
}

int main(){
    int n, m; cin >> n >> m;
    if(m%2){ cout << -1 << endl; return 0; }
    G.assign(n,vector<int>());
    dist.assign(n,-1);
    rep(i,m){
        int a, b; cin >> a >> b; a--; b--;
        G[a].pb(b); G[b].pb(a);
    }
    dfs(0,-1);
    return 0;
}