 #include<bits/stdc++.h>
 using namespace std;
 const int maxx=1e5+2;
 struct node
 {
     int x,y,id;
 };
 vector<node> res;
 struct S
 {
     int nex,id;
 };
 vector<S> b[maxx];
 int in[maxx],odd[maxx];
 int vis[maxx]; int sign[maxx];
 int f=0;
 void dfs(int x,int fa){
     vis[x]=1;
     if(odd[x]) f=1-f;
     for(auto t: b[x]){
         if(vis[t.nex]) continue;
         if(f){
            sign[t.id]=1-sign[t.id];
         }
         dfs(t.nex,x);
         if(f){
             sign[t.id]=1-sign[t.id];
         }
     }
 }
 int main(){
    int n,m; cin>>n>>m;
    for(int i=1;i<=m;i++){
        int x,y; scanf("%d%d",&x,&y);
        res.push_back(node{x,y,i});
        in[x]++;
        b[x].push_back(S{y,i});
        b[y].push_back(S{x,i});
    }
    if(m&1) return 0*puts("-1");
    for(int i=1;i<=n;i++){
        if(in[i]&1) odd[i]=1;
    }
    dfs(1,0);
    for(auto x:res){
        if(sign[x.id]){
            printf("%d %d\n",x.y,x.x);
        }else{
            printf("%d %d\n",x.x,x.y);
        }
    }
 }
### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxx=1e5+2;
 struct node
 {
     int x,y,id;
 };
 vector<node> res;
 struct S
 {
     int nex,id;
 };
 vector<S> b[maxx];
 int in[maxx],odd[maxx];
 int vis[maxx]; int sign[maxx];
 int f=0;
 void dfs(int x,int fa){
     vis[x]=1;
     if(odd[x]) f=1-f;
     for(auto t: b[x]){
         if(vis[t.nex]) continue;
         if(f){
            sign[t.id]=1-sign[t.id];
         }
         dfs(t.nex,x);
         if(f){
             sign[t.id]=1-sign[t.id];
         }
     }
 }
 int main(){
    int n,m; cin>>n>>m;
    for(int i=1;i<=m;i++){
        int x,y; scanf("%d%d",&x,&y);
        res.push_back(node{x,y,i});
        in[x]++;
        b[x].push_back(S{y,i});
        b[y].push_back(S{x,i});
    }
    if(m&1) return 0*puts("-1");
    for(int i=1;i<=n;i++){
        if(in[i]&1) odd[i]=1;
    }
    dfs(1,0);
    for(auto x:res){
        if(sign[x.id]){
            printf("%d %d\n",x.y,x.x);
        }else{
            printf("%d %d\n",x.x,x.y);
        }
    }
 }
```