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;
const int N=1e5+10;
struct node{
	int u,to,w;
};
node edge[N<<1];
bool b[N];
int dep[N],a[N],c[N],f[N],bb[N],head[N];
int k,n,m;
void add(int x,int y,int z) {
	edge[k].u=y; edge[k].to=head[x]; edge[k].w=z; head[x]=k++;
}
void dfs(int now,int fa,int d)
{
	b[now]=1; dep[now]=d;
	for (int i=head[now];i!=-1;i=edge[i].to) {
		int u=edge[i].u;
		if (!b[u]) dfs(u,now,d+1);
	}
	if (f[now]==1)
	for (int i=head[now];i!=-1;i=edge[i].to) {
		int u=edge[i].u;
		if (dep[u]<dep[now]) {
			f[now]^=1; f[u]^=1; c[edge[i].w]^=1; break;
		}
	}
}
int main() {
	scanf("%d%d",&n,&m);
		memset(head,-1,sizeof(head));
		for (int i=1;i<=m;i++) {
			scanf("%d%d",&a[i],&bb[i]); add(a[i],bb[i],i); add(bb[i],a[i],i);
			f[a[i]]^=1;
			c[i]=1;
	    }
	    for (int i=1;i<=n;i++) if (!b[i]){
	    	dfs(i,0,1);
		}
		for (int i=1;i<=n;i++) if (f[i]==1) {
			printf("%d\n",-1); return 0;
		}  
	    for (int i=1;i<=m;i++) if (c[i]) printf("%d %d\n",a[i],bb[i]); else printf("%d %d\n",bb[i],a[i]);
}