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;

const int N = 1e5 + 10;

vector<int> G[N], H[N];
int f[N];
int findSet (int x) { return x == f[x] ? x : f[x] = findSet(f[x]); }

void dfs (int u, int par) {
	for (auto & v : G[u]) {
		if (v == par) continue;
		dfs(v, u);
		if (H[v].size() & 1) {
			H[v].push_back(u);
		} else {
			H[u].push_back(v);
		}
	}
}

int main (void) {
	int n, m;
	scanf("%d%d", &n, &m);
	if (m & 1) {
		puts("-1");
		return 0;
	}
	for (int i = 1; i <= n; i++) {
		f[i] = i;
	}
	for (int i = 1, u, v; i <= m; i++) {
		scanf("%d%d", &u, &v);
		int uu = findSet(u);
		int vv = findSet(v);
		if (uu != vv) {
			G[v].push_back(u);
			G[u].push_back(v);
			f[vv] = uu;
		} else {
			H[u].push_back(v);
		}
	}
	dfs(1, 1);
	for (int i = 1; i <= n; i++) {
		for (auto & j : H[i]) {
			printf("%d %d\n", i, j);
		}
	}
}