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;

vector<int> e[100010];
int c[100010];
int used[100010];
map<pair<int, int>, bool> mm;
void dfs(int curr, int prev) {
    used[curr] = 1;
    for (auto &to: e[curr]) {
        if (used[to]) continue;
        mm[{curr, to}] = 1;
        c[curr]++;
    }
    for (auto &to: e[curr]) {
        if (used[to]) continue;
        dfs(to, curr);
        while (c[to] % 2) {
            mm[{curr, to}] = 0;
            c[to]--;
            mm[{to, curr}] = 1;
            c[curr]++;
        }
    }
    return;
}

int main() {
    int n, m; cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int a, b; cin >> a >> b;
        a--, b--;
        e[a].emplace_back(b); e[b].emplace_back(a);
    }
    dfs(0, -1);
    if (c[0] % 2) cout << -1 << endl;
    else {
        int cnt = 0;
        for (auto &ans: mm) {
            if (ans.second) {
                cnt++;
                cout << ans.first.first+1 << " " << ans.first.second+1 << endl;
            }
        }
        assert(cnt == m);
    }
    return 0;
}