#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+100;
int Next[N*2],Head[N],To[N*2],Val[N*2],cnt;
int tag[N],x[N],y[N],z[N];
int n,m;
bool vis[N];
void add(int u, int v, int w){
    ++cnt; Next[cnt] = Head[u]; 
    Head[u] = cnt; To[cnt] = v; Val[cnt] = w;
}
void dfs(int u,int fa, int now){
    vis[u] = 1;
    for (int i = Head[u]; i; i = Next[i]){
        int v = To[i];
        if (vis[To[i]]) continue;
        dfs(v,u,Val[i]);
    }
    if (tag[u]){
        tag[u] ^= 1; tag[fa]^=1;
        z[now] = -z[now];
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for (int i = 0; i < m; ++i){
        scanf("%d%d",&x[i],&y[i]); z[i] = 1;
        add(x[i],y[i],i); add(y[i],x[i],i);
        tag[x[i]]^=1;
    }
    if (m & 1){
        puts("-1"); return 0;
    }
    dfs(1,0,0);
    for (int i = 0; i < m; ++i){
        if (z[i] > 0) printf("%d %d\n",x[i],y[i]); else printf("%d %d\n",y[i],x[i]);
    }
    return 0;
}
### Prompt

Your task is to create 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 = 1e5+100;
int Next[N*2],Head[N],To[N*2],Val[N*2],cnt;
int tag[N],x[N],y[N],z[N];
int n,m;
bool vis[N];
void add(int u, int v, int w){
    ++cnt; Next[cnt] = Head[u]; 
    Head[u] = cnt; To[cnt] = v; Val[cnt] = w;
}
void dfs(int u,int fa, int now){
    vis[u] = 1;
    for (int i = Head[u]; i; i = Next[i]){
        int v = To[i];
        if (vis[To[i]]) continue;
        dfs(v,u,Val[i]);
    }
    if (tag[u]){
        tag[u] ^= 1; tag[fa]^=1;
        z[now] = -z[now];
    }
}
int main(){
    scanf("%d%d",&n,&m);
    for (int i = 0; i < m; ++i){
        scanf("%d%d",&x[i],&y[i]); z[i] = 1;
        add(x[i],y[i],i); add(y[i],x[i],i);
        tag[x[i]]^=1;
    }
    if (m & 1){
        puts("-1"); return 0;
    }
    dfs(1,0,0);
    for (int i = 0; i < m; ++i){
        if (z[i] > 0) printf("%d %d\n",x[i],y[i]); else printf("%d %d\n",y[i],x[i]);
    }
    return 0;
}
```