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 pb push_back
using namespace std;

int n,m,a,b,d[200200];
vector<int> G[200200];

void pr(int a, int b){ printf("%d %d\n", a, b); }
void dfs(int v, int p){
    d[v]++;
    for(int nv : G[v]){
        if(nv == p) continue;
        if(d[nv]){
            if(v < nv) pr(v,nv), d[v]++;
        }
        else dfs(nv,v);
    }
    if(p){
        if(d[v]%2) pr(p,v), d[p]++;
        else pr(v,p), d[v]++;
    }
}

int main(){
    scanf("%d %d", &n, &m);
    if(m%2) printf("%d\n", -1);
    else{
        while(m--){
            scanf("%d %d", &a, &b);
            G[a].pb(b); G[b].pb(a);
        }
    dfs(1,0);
    }
}