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 N = 100000 + 10;
vector<pair<int,int> > g[N];
int vis[N],d[N];
set<pair<int,int> > s;
void dfs(int u){
    vis[u] = 1;
    for(auto i : g[u]){
        int v = i.first;
        if(vis[v]==0) dfs(v), d[u] ^= d[v];
    }
}
void solve(int u){
    vis[u] = 1;
    for(auto i : g[u]){
        int v = i.first;
        if(vis[v]) continue;
        if(i.second==d[v]) cout<<u<<' '<<v<<endl;
        else cout<<v<<' '<<u<<endl;
        if(i.second==0) s.erase(make_pair(u,v));
        else s.erase(make_pair(v,u));
        solve(v);
    }
}
int main(){
    int n,m; cin>>n>>m;
    for(int i=1;i<=m;i++){
        int u,v; cin>>u>>v;
        g[u].push_back(make_pair(v,0));
        g[v].push_back(make_pair(u,1));
        s.insert(make_pair(u,v));
        d[u] ^= 1;
    }
    if(m&1) cout<<-1<<endl;
    else{
        dfs(1);
        memset(vis,0,sizeof(vis));
        solve(1);
        for(auto x : s){
            cout<<x.first<<' '<<x.second<<endl;
        }
    }
}