#include<bits/stdc++.h>
using namespace std;
typedef long long ll ;
const  int maxn=1e5+10;
int depth[maxn];
int out_put[maxn];
struct Node
{
    int x,y;
};
vector<int>s[maxn];
void dfs(int now,int pre)
{
    for(int i=0; i<s[now].size(); i++)
    {
        int x=s[now][i];
        if(x==pre)
        continue;
        if(depth[x]==0)
        {
            depth[x]=depth[now]+1;
            dfs(x,now);
        }
        else
        {
            if(depth[x]>depth[now])
            {
                printf("%d %d\n",now,x);
                out_put[now]++;
            }
        }
    }
    if(pre==0)
        return ;
    if(out_put[now]%2)
    {
        printf("%d %d\n",now,pre);
        out_put[now]++;
    }
    else
    {
        printf("%d %d\n",pre,now);
        out_put[pre]++;
    }


}
int main()
{
    int m,n;
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        int x,y;
        cin>>x>>y;
        s[x].push_back(y);
        s[y].push_back(x);
    }
    if(m%2)
    {
        puts("-1");
        return 0;
    }
    depth[1]=1;
    dfs(1,0);
    return 0;
}

### Prompt

In CPP, your task is to solve 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;
typedef long long ll ;
const  int maxn=1e5+10;
int depth[maxn];
int out_put[maxn];
struct Node
{
    int x,y;
};
vector<int>s[maxn];
void dfs(int now,int pre)
{
    for(int i=0; i<s[now].size(); i++)
    {
        int x=s[now][i];
        if(x==pre)
        continue;
        if(depth[x]==0)
        {
            depth[x]=depth[now]+1;
            dfs(x,now);
        }
        else
        {
            if(depth[x]>depth[now])
            {
                printf("%d %d\n",now,x);
                out_put[now]++;
            }
        }
    }
    if(pre==0)
        return ;
    if(out_put[now]%2)
    {
        printf("%d %d\n",now,pre);
        out_put[now]++;
    }
    else
    {
        printf("%d %d\n",pre,now);
        out_put[pre]++;
    }


}
int main()
{
    int m,n;
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        int x,y;
        cin>>x>>y;
        s[x].push_back(y);
        s[y].push_back(x);
    }
    if(m%2)
    {
        puts("-1");
        return 0;
    }
    depth[1]=1;
    dfs(1,0);
    return 0;
}

```