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>
#define Rint register int
using namespace std;
const int N = 100003;
int n, m, head[N], to[N << 1], nxt[N << 1];
inline void add(int a, int b){
	static int cnt = 0;
	to[++ cnt] = b; nxt[cnt] = head[a]; head[a] = cnt;
}
int dfn[N], tim;
inline int dfs(int x, int f){
	dfn[x] = ++ tim;
	int v = 0;
	for(Rint i = head[x];i;i = nxt[i]) if(to[i] != f){
		if(!dfn[to[i]]){
			int val = dfs(to[i], x);
			if(val) printf("%d %d\n", x, to[i]);
			else printf("%d %d\n", to[i], x);
			v ^= val;
		} else {
			if(dfn[to[i]] < dfn[x]) continue;
			printf("%d %d\n", x, to[i]);
			v ^= 1;
		}
	}
	return 1 - v;
}
int main(){
	scanf("%d%d", &n, &m);
	if(m & 1){puts("-1"); return 0;}
	for(Rint i = 1;i <= m;i ++){
		int a, b;
		scanf("%d%d", &a, &b);
		add(a, b); add(b, a);
	}
	dfs(1, 0);
}