#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;

int N, M;
vector<int> adj[101010];
int deg[101010], par[101010];
int id[101010], num=1;

priority_queue<pii> PQ;

void dfs(int u){
    vector<int> E;
    id[u] = num++;
    for (int v : adj[u]) if (!id[v]) dfs(v);
    for (int v : adj[u]) if (id[v] < id[u]) E.push_back(v);
    if (par[u] & 1){
        printf("%d %d\n", u, E.back());
        E.pop_back();
    }
    while (E.size() > 1){
        printf("%d %d\n", u, E.back());
        E.pop_back();
        printf("%d %d\n", u, E.back());
        E.pop_back();
    }
    if (E.size()) {
        printf("%d %d\n", E[0], u);
        par[E[0]] = !par[E[0]];
    }
}

int main(){
    scanf("%d %d", &N, &M);
    if (M & 1){
        puts("-1");
        return 0;
    }
    for (int i=1; i<=M; i++){
        int u, v;
        scanf("%d %d", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs(1);
    return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
typedef pair<int, int> pii;

int N, M;
vector<int> adj[101010];
int deg[101010], par[101010];
int id[101010], num=1;

priority_queue<pii> PQ;

void dfs(int u){
    vector<int> E;
    id[u] = num++;
    for (int v : adj[u]) if (!id[v]) dfs(v);
    for (int v : adj[u]) if (id[v] < id[u]) E.push_back(v);
    if (par[u] & 1){
        printf("%d %d\n", u, E.back());
        E.pop_back();
    }
    while (E.size() > 1){
        printf("%d %d\n", u, E.back());
        E.pop_back();
        printf("%d %d\n", u, E.back());
        E.pop_back();
    }
    if (E.size()) {
        printf("%d %d\n", E[0], u);
        par[E[0]] = !par[E[0]];
    }
}

int main(){
    scanf("%d %d", &N, &M);
    if (M & 1){
        puts("-1");
        return 0;
    }
    for (int i=1; i<=M; i++){
        int u, v;
        scanf("%d %d", &u, &v);
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    dfs(1);
    return 0;
}

```