#include <bits/stdc++.h>
using namespace std;

int t1,t2;

int n,m;
set<int> g[100005];
vector<int> ans[100005];
bool vis[100005];

void dfs(int u,int p)
{
    if (vis[u]) return;
    vis[u]=1;
    while (!g[u].empty()) {
        int v=*g[u].begin();
        g[u].erase(v);
        g[v].erase(u);
        if (vis[v]) {
            ans[u].push_back(v);
        }
        else {
            dfs(v,u);
        }
    }
    if (p) {
        if (ans[u].size()&1) ans[u].push_back(p);
        else ans[p].push_back(u);
    }
}

int main()
{
    #ifdef NOT_DMOJ
    freopen("data.txt","r",stdin);
    #endif // NOT_DMOJ
    scanf("%d%d",&n,&m);
    if (m&1) {
        printf("-1\n");
        return 0;
    }
    while (m--) {
        scanf("%d%d",&t1,&t2);
        g[t1].insert(t2);
        g[t2].insert(t1);
    }
    dfs(1,0);
    for (int i=1; i<=n; i++) {
        for (int j:ans[i]) {
            printf("%d %d\n",i,j);
        }
    }
}

### Prompt

Create a solution in cpp for 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;

int t1,t2;

int n,m;
set<int> g[100005];
vector<int> ans[100005];
bool vis[100005];

void dfs(int u,int p)
{
    if (vis[u]) return;
    vis[u]=1;
    while (!g[u].empty()) {
        int v=*g[u].begin();
        g[u].erase(v);
        g[v].erase(u);
        if (vis[v]) {
            ans[u].push_back(v);
        }
        else {
            dfs(v,u);
        }
    }
    if (p) {
        if (ans[u].size()&1) ans[u].push_back(p);
        else ans[p].push_back(u);
    }
}

int main()
{
    #ifdef NOT_DMOJ
    freopen("data.txt","r",stdin);
    #endif // NOT_DMOJ
    scanf("%d%d",&n,&m);
    if (m&1) {
        printf("-1\n");
        return 0;
    }
    while (m--) {
        scanf("%d%d",&t1,&t2);
        g[t1].insert(t2);
        g[t2].insert(t1);
    }
    dfs(1,0);
    for (int i=1; i<=n; i++) {
        for (int j:ans[i]) {
            printf("%d %d\n",i,j);
        }
    }
}

```