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<stdio.h>
#include<algorithm>
#include<vector>
#define fst first
#define snd second

using namespace std;

typedef pair<int, int> pii;
typedef long long ll;

const int MN = 1e5 + 1;
const int MM = 1e5 + 1;

int N, M;
vector<int> E[MN];
int outdegree[MN], visited[MN];

void connect(int a, int b) {
    outdegree[a] ^= 1;
    printf("%d %d\n", a+1, b+1);
}

void dfs(int n, int prev) {
    visited[n] = 1;
    for(int t: E[n]) {
        if (visited[t]) {
            if (t != prev && visited[t] == 1) {
                connect(n, t);
            }
        }
        else {
            dfs(t, n);
        }
    }
    if (prev != -1) {
        if (outdegree[n] == 1) {
            connect(n, prev);
        }
        else {
            connect(prev, n);
        }
    }
    visited[n] = 2;
}

int main() {
    scanf("%d%d", &N, &M);
    if (M&1) {
        return !puts("-1");
    }
    
    for(int i = 0; i < M; i++) {
        int a, b;
        scanf("%d%d", &a, &b); a--, b--;
        E[a].push_back(b), E[b].push_back(a);
    }
    dfs(0, -1);
    return 0;
}