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>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 1e5+100;
struct Edge{
    int u,v;
}edges[maxn];
int n,m;
vector<int> g[maxn];
int vis[maxn],d[maxn];

void dfs(int u){
    vis[u]=1; int v;
    for(int i=0;i<g[u].size();++i){
        Edge &e=edges[g[u][i]]; v=e.u^e.v^u;
        if(vis[v]) continue;
        dfs(v); d[u]^=d[v];
        if(d[v]==1) swap(e.u,e.v);
    }
}

int main(){
    scanf("%d %d",&n,&m);
    for(int i=0;i<m;++i){
        scanf("%d %d",&edges[i].u,&edges[i].v); d[edges[i].u]^=1;
        g[edges[i].u].push_back(i); g[edges[i].v].push_back(i);
    }
    dfs(1);
    if(d[1]) printf("-1\n");
    else{
        for(int i=0;i<m;++i) printf("%d %d\n",edges[i].u,edges[i].v);
    }
}