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

#define N 100100

int n, m, deg[N], vis[N];
vector <int> adj[N];
typedef pair <int, int> pii;
vector <pii> ans;

void dfs(int u, int pa = 0) {
    vis[u] = 1;
    for (auto v : adj[u]) {
        if (vis[v]) {
            if (v != pa && vis[v] == 1) {
                ans.emplace_back(pii(u, v));
                deg[u] ++;
            }
            continue;
        }
        dfs(v, u);
    }
    if (pa) {
        if (deg[u] & 1) ans.emplace_back(pii(u, pa)), deg[u] ++;
        else ans.emplace_back(pii(pa, u)), deg[pa] ++;
    }
    vis[u] = 2;
}

int main() {
    scanf("%d %d", &n, &m);
    while (m --) {
        int u, v;
        scanf("%d %d", &u, &v);
        adj[u].push_back(v), adj[v].push_back(u);
    }
    for (int i = 1; i <= n; i ++) if (!vis[i]) {
        dfs(i); if (deg[i] & 1) return puts("-1"), 0;
    }
    for (auto p : ans) printf("%d %d\n", p.first, p.second);

    return 0;
}

### 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;

#define N 100100

int n, m, deg[N], vis[N];
vector <int> adj[N];
typedef pair <int, int> pii;
vector <pii> ans;

void dfs(int u, int pa = 0) {
    vis[u] = 1;
    for (auto v : adj[u]) {
        if (vis[v]) {
            if (v != pa && vis[v] == 1) {
                ans.emplace_back(pii(u, v));
                deg[u] ++;
            }
            continue;
        }
        dfs(v, u);
    }
    if (pa) {
        if (deg[u] & 1) ans.emplace_back(pii(u, pa)), deg[u] ++;
        else ans.emplace_back(pii(pa, u)), deg[pa] ++;
    }
    vis[u] = 2;
}

int main() {
    scanf("%d %d", &n, &m);
    while (m --) {
        int u, v;
        scanf("%d %d", &u, &v);
        adj[u].push_back(v), adj[v].push_back(u);
    }
    for (int i = 1; i <= n; i ++) if (!vis[i]) {
        dfs(i); if (deg[i] & 1) return puts("-1"), 0;
    }
    for (auto p : ans) printf("%d %d\n", p.first, p.second);

    return 0;
}

```