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;

vector<int> edge[100005];
bool vis[100005], A[100005], B[100005];

void dfs(int x, int p) {
	vis[x] = 1;
	for (auto i : edge[x]) {
		if (vis[i]) {
			if (i != p && x < i) A[x] ^= 1;
			continue;
		}
		A[x] ^= 1;
		dfs(i, x);
	}
}

bool f(int x, int p) {
	bool ret = B[x];
	vis[x] = 1;
	for (auto i : edge[x]) {
		if (vis[i]) {
			if(i != p && x < i) printf("%d %d\n",x, i);
			continue;
		}
		bool t = f(i, x);
		if (t) printf("%d %d\n",i, x);
		else printf("%d %d\n", x, i);
		ret ^= t;
	}
	return ret;
}

int main() {
 	ios_base::sync_with_stdio(0), cin.tie(0);
 	int N, M; cin >> N >> M;
 	for (int i = 0; i < M; ++i) {
 		int x, y; cin >> x >> y;
 		edge[x].push_back(y);
 		edge[y].push_back(x);
 	}
 	if (M & 1) return !printf("-1");
 	dfs(1, 1);
 	vector<int> vec;
 	for (int i = 1; i <= N; ++i) if(A[i]) vec.push_back(i);
 	for (int i = 0; i < vec.size(); i += 2) {
 		B[vec[i]] ^= 1;
 		B[vec[i + 1]] ^= 1;
 	}
 	memset(vis, 0, sizeof(vis));
 	f(1, 1);
	return 0;
}