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;

int n, m;
int vis[100005];
int fr[100005], to[100005];
vector <int> adj[100005];
int a[100005];
int swp[100005];
int dad[100005], id[100005];

void dfs(int u) {
    swp[u] = a[u];
    for (auto x: adj[u]) {
        int v = u ^ fr[x] ^ to[x];
        if (vis[v]) continue;
        vis[v] = 1;
        dad[v] = u;
        id[v] = x;
        dfs(v);
        swp[u] ^= swp[v];
    }
}

int main(void) {
    ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        cin >> fr[i] >> to[i];
        a[fr[i]] ^= 1;
        adj[fr[i]].push_back(i);
        adj[to[i]].push_back(i);
    }
    if (m & 1) return cout << "-1" << endl, 0;
    vis[1] = 1;
    dfs(1);
    for (int i = 1; i <= n; ++i) if (swp[i]) {
        int x = id[i];
        swap(fr[x], to[x]);
    }
    for (int i = 1; i <= m; ++i) cout << fr[i] << ' ' << to[i] << endl;
    return 0;
}