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

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

const int N = 1e5 + 5;
int fa[N], deg[N];
vector<int> G[N];

int find(int x){ return x == fa[x] ? x : fa[x] = find(fa[x]); }

void dfs(int u, int fa)
{
	for (int v : G[u]) 
		if (v != fa) dfs(v, u);
	
	if (fa)
	{
		if (deg[u] & 1)
			printf("%d %d\n", u, fa);
		else
			printf("%d %d\n", fa, u), ++deg[fa];
	}
}

int main()
{
	int n = read(), m = read();
	if (m & 1) return puts("-1"), 0;
	
	for (int i = 1; i <= n; ++i) fa[i] = i;
	for (int i = 1; i <= m; ++i)
	{
		int u = read(), v = read();
		int fu = find(u), fv = find(v);
		if (fu != fv)
		{
			fa[fu] = fv;
			G[u].push_back(v);
			G[v].push_back(u);
		}
		else
		{
			printf("%d %d\n", u, v);
			++deg[u];
		}
	}
	
	dfs(1, 0);
	
	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<bits/stdc++.h>
using namespace std;

inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	for (; ch < '0' || ch > '9'; ch = getchar()) if (ch == '-') f = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

const int N = 1e5 + 5;
int fa[N], deg[N];
vector<int> G[N];

int find(int x){ return x == fa[x] ? x : fa[x] = find(fa[x]); }

void dfs(int u, int fa)
{
	for (int v : G[u]) 
		if (v != fa) dfs(v, u);
	
	if (fa)
	{
		if (deg[u] & 1)
			printf("%d %d\n", u, fa);
		else
			printf("%d %d\n", fa, u), ++deg[fa];
	}
}

int main()
{
	int n = read(), m = read();
	if (m & 1) return puts("-1"), 0;
	
	for (int i = 1; i <= n; ++i) fa[i] = i;
	for (int i = 1; i <= m; ++i)
	{
		int u = read(), v = read();
		int fu = find(u), fv = find(v);
		if (fu != fv)
		{
			fa[fu] = fv;
			G[u].push_back(v);
			G[v].push_back(u);
		}
		else
		{
			printf("%d %d\n", u, v);
			++deg[u];
		}
	}
	
	dfs(1, 0);
	
	return 0;
}

```