#include <cstdio>
#include <vector>

using namespace std;

int c[100000];
int d[100000];
vector <int> v[100000];

void dfs(int x, int z) {
    int i;
    
    d[x] = z;
    
    for (i = 0; i < v[x].size(); i++) {
        int y = v[x][i];
        
        if (d[y] == 0) {
            dfs(y, z + 1);
            
            if (c[y] % 2 == 0) {
                c[x]++;
                printf("%d %d\n", x + 1, y + 1);
            } else {
                c[y]++;
                printf("%d %d\n", y + 1, x + 1);
            }
        } else if (d[y] > d[x]) {
            c[x]++;
            printf("%d %d\n", x + 1, y + 1);
        }
    }
}

int main() {
    int n, m, i;
    
    scanf("%d %d", &n, &m);
    
    for (i = 0; i < m; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        x--;
        y--;
        
        v[x].push_back(y);
        v[y].push_back(x);
    }
    
    if (m % 2 == 1) {
        puts("-1");
        
        return 0;
    }
    
    dfs(0, 1);
    
    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 <cstdio>
#include <vector>

using namespace std;

int c[100000];
int d[100000];
vector <int> v[100000];

void dfs(int x, int z) {
    int i;
    
    d[x] = z;
    
    for (i = 0; i < v[x].size(); i++) {
        int y = v[x][i];
        
        if (d[y] == 0) {
            dfs(y, z + 1);
            
            if (c[y] % 2 == 0) {
                c[x]++;
                printf("%d %d\n", x + 1, y + 1);
            } else {
                c[y]++;
                printf("%d %d\n", y + 1, x + 1);
            }
        } else if (d[y] > d[x]) {
            c[x]++;
            printf("%d %d\n", x + 1, y + 1);
        }
    }
}

int main() {
    int n, m, i;
    
    scanf("%d %d", &n, &m);
    
    for (i = 0; i < m; i++) {
        int x, y;
        
        scanf("%d %d", &x, &y);
        
        x--;
        y--;
        
        v[x].push_back(y);
        v[y].push_back(x);
    }
    
    if (m % 2 == 1) {
        puts("-1");
        
        return 0;
    }
    
    dfs(0, 1);
    
    return 0;
}

```