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 MAXNUM 111111
vector<int> e[MAXNUM],te[MAXNUM];
int isok[MAXNUM],dis[MAXNUM];
void dfs(int now,int p,int d)
{
    dis[now]=d;
    for(int k:e[now])if(k!=p)
    {
        if(!dis[k])isok[now]^=1,te[now].push_back(k),dfs(k,now,d+1);
        else if(dis[k]<d)isok[now]^=1,printf("%d %d\n",now,k);
    }
}
void dfs2(int now,int p)
{
    for(int k:te[now])
        dfs2(k,now),isok[now]^=isok[k];
    if(p&&!isok[now])printf("%d %d\n",p,now);
    else if(p)printf("%d %d\n",now,p);
}
int main()
{
    int n,m;scanf("%d%d",&n,&m);
    if(m&1)printf("-1\n");
    else{
        int a,b;
        for(int i=1;i<=m;i++)
            scanf("%d%d",&a,&b),e[a].push_back(b),e[b].push_back(a);
    }
    dfs(1,0,1),dfs2(1,0);
}