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:
#if 0
date +%Y.%m.%d

题解先咕着。
#endif
#include <cstdio>
#include <vector>
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef std::pair<int, int> par;
inline int input() { int x; scanf("%d", &x); return x; }

const int maxn = 100005;
std::vector<par> G[maxn];
bool a[maxn], vis[maxn];
par edge[maxn];
bool ans[maxn];

void dfs(int u) {
	vis[u] = true;
	for(par p : G[u])
		if(not vis[p.first]) {
			dfs(p.first);
			if(a[p.first]) {
				a[u] ^= 1;
				a[p.first] ^= 1;
				ans[p.second] ^= 1;
			}
		}
}

int main() {
	int n = input(), m = input();
	for(int i = 1; i <= m; i ++) {
		int u = input(), v = input();
		G[u].push_back(par(v, i));
		G[v].push_back(par(u, i));
		edge[i] = par(u, v);
		a[u] ^= 1;
	}
	dfs(1);
	if(a[1])
		puts("-1");
	else {
		for(int i = 1; i <= m; i ++) {
			if(ans[i]) std::swap(edge[i].first, edge[i].second);
			printf("%d %d\n", edge[i].first, edge[i].second);
		}
	}
}