#include<bits/stdc++.h>
using namespace std;

#define N 300005

int n,m,x,y,d[N],a[N],b[N],k;
vector<int>g[N];
int vis[N];

bool dfs(int u,int fa)
{
    int s=0;
    vis[u]=1,d[u]=d[fa]+1;
    for(auto v:g[u])
    {
        if(!vis[v])
            s^=dfs(v,u);
        else if(d[v]>d[u])
        {
            a[++k]=u,b[k]=v;
            s^=1;
        }
    }
    if(s)
        a[++k]=u,b[k]=fa;
    else
        a[++k]=fa,b[k]=u;
    return s^1;
}
int main()
{
   scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    if(!dfs(1,0))
        puts("-1");
    else
    {
        for(int i=1;i<=m;i++)
            printf("%d %d\n",a[i],b[i]);
    }
    return 0;
}

### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

#define N 300005

int n,m,x,y,d[N],a[N],b[N],k;
vector<int>g[N];
int vis[N];

bool dfs(int u,int fa)
{
    int s=0;
    vis[u]=1,d[u]=d[fa]+1;
    for(auto v:g[u])
    {
        if(!vis[v])
            s^=dfs(v,u);
        else if(d[v]>d[u])
        {
            a[++k]=u,b[k]=v;
            s^=1;
        }
    }
    if(s)
        a[++k]=u,b[k]=fa;
    else
        a[++k]=fa,b[k]=u;
    return s^1;
}
int main()
{
   scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    if(!dfs(1,0))
        puts("-1");
    else
    {
        for(int i=1;i<=m;i++)
            printf("%d %d\n",a[i],b[i]);
    }
    return 0;
}

```