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 maxn = 1e5 + 10;

int vis[maxn], dep[maxn];

vector<int> edge[maxn];
vector<pair<int, int> > ans;

bool dfs(int u, int pre)
{
    int s = 0;
    vis[u] = 1, dep[u] = dep[pre] + 1;
    for(auto v : edge[u])
    {
        if(!vis[v]) s ^= dfs(v, u);
        else if(dep[v] > dep[u])
        {
            ans.push_back({u, v});
            s ^= 1;
        }
    }
    if(s && pre) ans.push_back({u, pre});
    else if(pre) ans.push_back({pre, u});
    return s ^ 1;
}

int main()
{
    int n, m, u, v;
    scanf("%d%d", &n, &m);
    while(m --)
    {
        scanf("%d%d", &u, &v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    if(!dfs(1, 0)) return puts("-1"), 0;
    for(auto i : ans) printf("%d %d\n", i.first, i.second);
    return 0;
}