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

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>

const int MN = 100005;

int N, M;
std::vector<int> G[MN];
int vis[MN], dep[MN], col[MN];

void DFS(int u, int f) {
	vis[u] = 1, dep[u] = dep[f] + 1;
	for (auto v : G[u]) {
		if (v == f) continue;
		if (vis[v]) {
			if (dep[v] < dep[u])
				col[u] ^= 1,
				printf("%d %d\n", u, v);
		}
		else DFS(v, u);
	}
	if (col[u]) printf("%d %d\n", u, f);
	else if (f) col[f] ^= 1, printf("%d %d\n", f, u);
}

int main() {
	int x, y;
	scanf("%d%d", &N, &M);
	if (M & 1) return puts("-1"), 0;
	for (int i = 1; i <= M; ++i)
		scanf("%d%d", &x, &y),
		G[x].push_back(y),
		G[y].push_back(x);
	DFS(1, 0);
	return 0;
}