#include <bits/stdc++.h>
using namespace std;
const int N=100005;
vector<pair<int,int> >G[N];
int n,m,a[N],b[N],deg[N],f[N];
int Find(int x)
{
    if(x==f[x]) return x;
    return f[x]=Find(f[x]);
}
void dfs(int u,int f)
{
    for(auto p:G[u])
    {
        int v=p.first,id=p.second;
        if(v==f) continue;
        dfs(v,u);
        if(deg[v]%2==0)
            deg[u]++,a[id]=u,b[id]=v;
        else
            deg[v]++,a[id]=v,b[id]=u;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    if(m%2) {printf("-1\n");return 0;}
    for(int i=1;i<=n;i++) f[i]=i;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        if(Find(a[i])!=Find(b[i]))
        {
            G[a[i]].push_back({b[i],i});
            G[b[i]].push_back({a[i],i});
            f[Find(a[i])]=Find(b[i]);
        }
        else deg[a[i]]++;
    }
    dfs(1,0);
    for(int i=0;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;
const int N=100005;
vector<pair<int,int> >G[N];
int n,m,a[N],b[N],deg[N],f[N];
int Find(int x)
{
    if(x==f[x]) return x;
    return f[x]=Find(f[x]);
}
void dfs(int u,int f)
{
    for(auto p:G[u])
    {
        int v=p.first,id=p.second;
        if(v==f) continue;
        dfs(v,u);
        if(deg[v]%2==0)
            deg[u]++,a[id]=u,b[id]=v;
        else
            deg[v]++,a[id]=v,b[id]=u;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    if(m%2) {printf("-1\n");return 0;}
    for(int i=1;i<=n;i++) f[i]=i;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        if(Find(a[i])!=Find(b[i]))
        {
            G[a[i]].push_back({b[i],i});
            G[b[i]].push_back({a[i],i});
            f[Find(a[i])]=Find(b[i]);
        }
        else deg[a[i]]++;
    }
    dfs(1,0);
    for(int i=0;i<m;i++)
        printf("%d %d\n",a[i],b[i]);
    return 0;
}
```