#include <cstdio>

#define forto(_) for (int e = back[_], v = E[e].to; e; \
		v = E[e = E[e].next].to)

const int kMaxn = 1e5 + 10;

struct Edge {
	int to, next;
} E[kMaxn << 1];

int n, m, tote, dfs_clock;
int back[kMaxn], dfn[kMaxn], odeg[kMaxn];

inline void AddEdge(int u, int v) {
	E[++tote] = (Edge){v, back[u]}, back[u] = tote;
	E[++tote] = (Edge){u, back[v]}, back[v] = tote;
}
void Dfs(int, int);

int main() {
	scanf("%d%d", &n, &m);
	if (m & 1) {
		puts("-1");
		return 0;
	}
	for (int i = 0, u, v; i < m; i++) {
		scanf("%d%d", &u, &v);
		AddEdge(u, v);
	}

	Dfs(1, 0);
	return 0;
}

void Dfs(int u, int fa) {
	dfn[u] = ++dfs_clock;
	forto(u) {
		if (v == fa) continue;
		if (!dfn[v]) Dfs(v, u);
		else if (dfn[v] < dfn[u]) {
			++odeg[u];
			printf("%d %d\n", u, v);
		}
	}
	if (u == 1) return;
	if (odeg[u] & 1) {
		++odeg[u];
		printf("%d %d\n", u, fa);
	} else {
		++odeg[fa];
		printf("%d %d\n", fa, u);
	}
}

### Prompt

Generate a Cpp solution to 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 <cstdio>

#define forto(_) for (int e = back[_], v = E[e].to; e; \
		v = E[e = E[e].next].to)

const int kMaxn = 1e5 + 10;

struct Edge {
	int to, next;
} E[kMaxn << 1];

int n, m, tote, dfs_clock;
int back[kMaxn], dfn[kMaxn], odeg[kMaxn];

inline void AddEdge(int u, int v) {
	E[++tote] = (Edge){v, back[u]}, back[u] = tote;
	E[++tote] = (Edge){u, back[v]}, back[v] = tote;
}
void Dfs(int, int);

int main() {
	scanf("%d%d", &n, &m);
	if (m & 1) {
		puts("-1");
		return 0;
	}
	for (int i = 0, u, v; i < m; i++) {
		scanf("%d%d", &u, &v);
		AddEdge(u, v);
	}

	Dfs(1, 0);
	return 0;
}

void Dfs(int u, int fa) {
	dfn[u] = ++dfs_clock;
	forto(u) {
		if (v == fa) continue;
		if (!dfn[v]) Dfs(v, u);
		else if (dfn[v] < dfn[u]) {
			++odeg[u];
			printf("%d %d\n", u, v);
		}
	}
	if (u == 1) return;
	if (odeg[u] & 1) {
		++odeg[u];
		printf("%d %d\n", u, fa);
	} else {
		++odeg[fa];
		printf("%d %d\n", fa, u);
	}
}

```