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;
#define maxn 100009
vector <int> ed[maxn];
int fi[maxn]={0};
int du[maxn]={0};
int n,m,tt=1;
void dfs(int x,int fa) {
    int a,b,c,i,j;
    fi[x]=tt++;
    for(a=0;a<ed[x].size();a++) 
    if(ed[x][a]!=fa) {
        if(fi[ed[x][a]]&&fi[ed[x][a]]<fi[x]) {printf("%d %d\n",x,ed[x][a]);du[x]^=1;continue;}
        else if(!fi[ed[x][a]]) dfs(ed[x][a],x);
    }
    if(x==1) return;
    if(du[x]) {printf("%d %d\n",x,fa);du[x]^=1;}
    else {printf("%d %d\n",fa,x);du[fa]^=1;}
}
int main(){
    scanf("%d%d",&n,&m);
    if(m%2) {printf("-1\n");return 0;}
    int a,b,c,i,j;
    for(a=1;a<=m;a++) {
        scanf("%d%d",&i,&j);
        ed[i].push_back(j);
      	ed[j].push_back(i);
    }
    dfs(1,0);
    return 0;
}