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>
#include <algorithm>
using namespace std;

const int Maxn = 1e5;

int N, M;
vector<int> G[Maxn + 5];
int dep[Maxn + 5];
vector<pair<int, int> > edges;

int DFS(int u, int fa) {
	dep[u] = dep[fa] + 1;
	int deg = 0;
	for(int i = 0; i < (int)G[u].size(); i++) {
		int v = G[u][i];
		if(v == fa) continue;
		if(dep[v]) {
			if(dep[v] < dep[u])
				deg++, edges.push_back(make_pair(u, v));
			continue;
		}
		int degv = DFS(v, u);
		if(degv & 1) edges.push_back(make_pair(v, u));
		else edges.push_back(make_pair(u, v)), deg++;
	}
	return deg;
}

int main() {
#ifdef LOACL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	scanf("%d %d", &N, &M);
	for(int i = 1; i <= M; i++) {
		int u, v;
		scanf("%d %d", &u, &v);
		G[u].push_back(v), G[v].push_back(u);
	}
	dep[1] = 1;
	int res = DFS(1, 0);
	if(res & 1) {
		puts("-1");
		return 0;
	}
	for(int i = 0; i < (int)edges.size(); i++)
		printf("%d %d\n", edges[i].first, edges[i].second);
	return 0;
}