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 <bits/stdc++.h>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

struct edge {
	int to;
	bool rev;
};

int N, M, a, b;
bool vis[100001];
vector<edge>v[100001];

bool dfs(int x,int bf) {
	vis[x] = true;
	bool c = false;
	for (edge e : v[x]) {
		if (!e.rev)c ^= true;
	}
	for (int i = 0; i < v[x].size(); ++i) {
		if (!vis[v[x][i].to]) {
			if (dfs(v[x][i].to, x)) {
				c ^= true;
				v[x][i].rev ^= true;
			}
		}
	}
	if (c) {
		for (edge &e : v[x]) {
			if (e.to == bf) {
				e.rev ^= true;
				break;
			}
		}
		return true;
	}
	return false;
}


int main() {
	cin >> N >> M;
	if (M & 1) {
		cout << -1 << endl;
		return 0;
	}
	for (int i = 0; i < M; ++i) {
		cin >> a >> b;
		v[a].push_back({ b, false });
		v[b].push_back({ a,true });
	}
	dfs(1, -1);
	for (int i = 1; i <= N; ++i) {
		for (edge e : v[i]) {
			if (!e.rev) {
				cout << i << sp << e.to << endl;
			}
		}
	}
}