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>
using namespace std;
typedef bool boolean;

const int N = 1e5 + 5;

int n, m;
int uf[N];
int d[N];
vector<int> G[N];

int find(int x) {
	return uf[x] == x ? x : (uf[x] = find(uf[x]));
}

void dfs(int p, int fa) {
	for (auto e : G[p]) {
		if (e ^ fa) {
			dfs(e, p);
		}
	}
	if (!fa)
		return;
	if (d[p]) {
		d[p] ^= 1;
		printf("%d %d\n", p, fa);
	} else {
		d[fa] ^= 1;
		printf("%d %d\n", fa, p);
	}
}

int main() {
	scanf("%d%d", &n, &m);
	if (m & 1) {
		puts("-1");
		return 0;
	}
	for (int i = 1; i <= n; i++)
		uf[i] = i;
	for (int i = 1, u, v; i <= m; i++) {
		scanf("%d%d", &u, &v);
		if (find(u) ^ find(v)) {
			uf[find(u)] = find(v);
			G[u].push_back(v);
			G[v].push_back(u);
		} else {
			printf("%d %d\n", u, v);
			d[u] ^= 1;
		}
	}
	dfs(1, 0);
	assert(!d[1]);
	return 0;
}