#include <iostream>
#include <vector>
using namespace std;
#define N 300000
 
int n,m,x,y,xb,p[N],q[N],dep[N];
bool vis[N];
vector<int> g[N];
 
bool dfs(int u, int fa) {
	int s = 0;
    vis[u] = 1;
    dep[u] = dep[fa] + 1;
	for ( auto v:g[u] ) {
		if ( !vis[v] ) {
			s ^= dfs(v, u);
		} else {
			if ( dep[v] > dep[u] ) {
				p[++xb] = u;
                q[xb] = v;
				s ^= 1;
			}
        }
    }
	if (s) {
        p[++xb] = u;
        q[xb] = fa;
    } else {
        p[++xb] = fa;
        q[xb] = u;
    }
	return s^1;
}

int main() {
	scanf("%d%d", &n, &m);
    //cin >> n >> m;
	for (int i=1; i<=m; ++i) {
		scanf("%d%d", &x, &y);
        //cin >> x >> y;
		g[x].push_back(y);
        g[y].push_back(x);
	}
	if ( !dfs(1,0) ) {
        puts("-1");
        //cout << "-1" << endl;
        return 0;
    }
	for (int i=1; i<=m; ++i) {
        printf("%d %d\n", p[i], q[i]);
        //cout << p[i] << ' ' << q[i] << endl;
    }

	return 0;
}

### 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 <iostream>
#include <vector>
using namespace std;
#define N 300000
 
int n,m,x,y,xb,p[N],q[N],dep[N];
bool vis[N];
vector<int> g[N];
 
bool dfs(int u, int fa) {
	int s = 0;
    vis[u] = 1;
    dep[u] = dep[fa] + 1;
	for ( auto v:g[u] ) {
		if ( !vis[v] ) {
			s ^= dfs(v, u);
		} else {
			if ( dep[v] > dep[u] ) {
				p[++xb] = u;
                q[xb] = v;
				s ^= 1;
			}
        }
    }
	if (s) {
        p[++xb] = u;
        q[xb] = fa;
    } else {
        p[++xb] = fa;
        q[xb] = u;
    }
	return s^1;
}

int main() {
	scanf("%d%d", &n, &m);
    //cin >> n >> m;
	for (int i=1; i<=m; ++i) {
		scanf("%d%d", &x, &y);
        //cin >> x >> y;
		g[x].push_back(y);
        g[y].push_back(x);
	}
	if ( !dfs(1,0) ) {
        puts("-1");
        //cout << "-1" << endl;
        return 0;
    }
	for (int i=1; i<=m; ++i) {
        printf("%d %d\n", p[i], q[i]);
        //cout << p[i] << ' ' << q[i] << endl;
    }

	return 0;
}

```