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 MAXN = 1.1e5;
const int MAXM = 1.1e5;
int N, M;
pair<int, int> E[MAXM];

vector<pair<int, int>> adj[MAXN];
int outDeg[MAXN]; // just parity

bool vis[MAXN];
void dfs(int cur) {
	assert(!vis[cur]);
	vis[cur] = true;
	for (auto it : adj[cur]) {
		int nxt = it.first;
		int e = it.second;
		if (vis[nxt]) continue;
		dfs(nxt);
		if (outDeg[nxt]) {
			swap(E[e].first, E[e].second);
			outDeg[nxt] ^= 1;
			outDeg[cur] ^= 1;
		}
		assert(outDeg[nxt] == 0);
	}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N >> M;
	for (int e = 0; e < M; e++) {
		cin >> E[e].first >> E[e].second;
		E[e].first --, E[e].second --;
	}

	if (M % 2 == 1) {
		cout << -1 << '\n';
		exit(0);
	}

	for (int e = 0; e < M; e++) {
		int u, v; tie(u, v) = E[e];
		adj[u].emplace_back(v, e);
		adj[v].emplace_back(u, e);
		outDeg[u] ^= 1;
	}

	dfs(0);
	for (int e = 0; e < M; e++) {
		cout << E[e].first+1 << ' ' << E[e].second+1 << '\n';
	}

	return 0;
}