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<algorithm>
#include<vector>
#define N_ 101000
using namespace std;
int n, m, Num[N_], cnt, Out[N_];
vector<int>E[N_], L[N_];
void DFS(int a, int pp) {
	Num[a] = ++cnt;
	for (int i = 0; i < E[a].size(); i++) {
		int x = E[a][i];
		if (Num[x])continue;
		DFS(x, a);
	}
	for (int i = 0; i < E[a].size(); i++) {
		int x = E[a][i];
		if (Num[x] < Num[a] && x != pp) {
			Out[a]++;
			printf("%d %d\n", a, x);
		}
	}
	if (pp) {
		if (Out[a] & 1) {
			Out[a]++;
			printf("%d %d\n", a, pp);
		}
		else {
			Out[pp]++;
			printf("%d %d\n", pp, a);
		}
	}
}
int main() {
	int i, a, b;
	scanf("%d%d", &n, &m);
	for (i = 1; i <= m; i++) {
		scanf("%d%d", &a, &b);
		E[a].push_back(b);
		E[b].push_back(a);
		L[a].push_back(i);
		L[b].push_back(i);
	}
	if (m % 2) {
		puts("-1");
		return 0;
	}
	DFS(1, 0);
}