#include "bits/stdc++.h"

using namespace std;

const int N = 1e5 + 5;
vector< pair< int, int>> conn[N];
int a[N], b[N];
bitset< N> seen;
int out[N], direction[N]; // 0 == a -> b, 1 == b -> a

void dfs(int u, int prt) {
    seen[u] = true;

    int index = -1;
    for (const auto& p : conn[u]){
        if (p.first == prt) {
            index = p.second;
            continue;
        }
        if (seen[p.first]) continue;
        dfs(p.first, u);
    }

    if (out[u] & 1) {
        out[a[index]] --;
        out[b[index]] ++;
        direction[index] = 1;
    }
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);

    if (m & 1) return !printf("-1");
    
    for (int i = 0; i < m; ++i) {
        scanf("%d %d", &a[i], &b[i]);
        conn[a[i]].emplace_back(b[i], i);
        conn[b[i]].emplace_back(a[i], i);
        out[a[i]] ++;
    }

    dfs(1, 1);
    for (int i = 0; i < m; ++i) {
        if (direction[i] == 0) printf("%d %d\n", a[i], b[i]);
        else printf("%d %d\n", b[i], a[i]);
    }
}
### 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;

const int N = 1e5 + 5;
vector< pair< int, int>> conn[N];
int a[N], b[N];
bitset< N> seen;
int out[N], direction[N]; // 0 == a -> b, 1 == b -> a

void dfs(int u, int prt) {
    seen[u] = true;

    int index = -1;
    for (const auto& p : conn[u]){
        if (p.first == prt) {
            index = p.second;
            continue;
        }
        if (seen[p.first]) continue;
        dfs(p.first, u);
    }

    if (out[u] & 1) {
        out[a[index]] --;
        out[b[index]] ++;
        direction[index] = 1;
    }
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);

    if (m & 1) return !printf("-1");
    
    for (int i = 0; i < m; ++i) {
        scanf("%d %d", &a[i], &b[i]);
        conn[a[i]].emplace_back(b[i], i);
        conn[b[i]].emplace_back(a[i], i);
        out[a[i]] ++;
    }

    dfs(1, 1);
    for (int i = 0; i < m; ++i) {
        if (direction[i] == 0) printf("%d %d\n", a[i], b[i]);
        else printf("%d %d\n", b[i], a[i]);
    }
}
```