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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int read(){
    int a=0,b=getchar(),c=1;
    while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
    while(isdigit(b))a=a*10+b-'0',b=getchar();
    return a*c;
}
void print(int x){
    char y=x%10+'0';
    if(x>=10)print(x/10);
    putchar(y);
}
int n,m,vis[100005];
vector<int> g[100005],ans[100005];
bool dfs(int x,int f){
	vis[x]=1;
	for(int i=0;i<g[x].size();i++)
		if(!vis[g[x][i]]){
			if(dfs(g[x][i],x))ans[g[x][i]].push_back(x);
			else ans[x].push_back(g[x][i]);
		}
		else if(g[x][i]!=f and vis[g[x][i]]==1)ans[x].push_back(g[x][i]);
	vis[x]=2;
	return ans[x].size()%2;
}
int main(){
	n=read(),m=read();
	for(int i=0;i<m;i++){
		int x=read(),y=read();
		g[x].push_back(y);
		g[y].push_back(x);
	}
	if(m%2){
		puts("-1");
		return 0;
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		for(int j=0;j<ans[i].size();j++)
			print(i),putchar(' '),print(ans[i][j]),puts("");
	return 0;
}