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<cstdio>
using namespace std;
const int maxn=100005,maxe=200005;
int n,m,p,ans[maxe][2];
int e=1,lnk[maxn],nxt[maxe],son[maxe];
bool vis[maxn],flg[maxe],F[maxn];
inline int read(){
	int ret=0;bool f=0;char ch=getchar();
	while(ch>'9'||ch<'0') f^=ch=='-',ch=getchar();
	while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=getchar();
	return f?-ret:ret;
}
void add_e(int x,int y){son[++e]=y;nxt[e]=lnk[x];lnk[x]=e;}
void DFS1(int x){
	vis[x]=1;
	for (int i=lnk[x];i;i=nxt[i]){
		if(vis[son[i]]) continue;
		flg[i]=1;DFS1(son[i]);
	}
}
void DFS2(int x){
	for (int i=lnk[x];i;i=nxt[i])
	if(flg[i]){
		DFS2(son[i]);
		if(F[son[i]]) ans[++p][0]=son[i],ans[p][1]=x;
		else F[x]^=1,ans[++p][0]=x,ans[p][1]=son[i];
	}
}
int main(){
	n=read(),m=read();
	if(m&1){puts("-1");return 0;}
	for (int i=1;i<=m;i++){
		int x=read(),y=read();
		add_e(x,y),add_e(y,x);
	}
	DFS1(1);
	for (int i=1;i<=n;i++)
	for (int j=lnk[i];j;j=nxt[j])
	if((j&1)&&!flg[j]&&!flg[j^1])
	ans[++p][0]=i,ans[p][1]=son[j],F[i]^=1;
	DFS2(1);
	for (int i=1;i<=m;i++)
	printf("%d %d\n",ans[i][0],ans[i][1]);
	return 0;
}