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 = 2e5 + 7;
int n, m, d[maxn], fa[maxn];

int find(int x) {
	return x == fa[x]? x : fa[x] = find(fa[x]);
}

int fir[maxn], nxt[maxn], to[maxn], e = 0;
void add(int x, int y) {
	to[++e] = y; nxt[e] = fir[x]; fir[x] = e;
	to[++e] = x; nxt[e] = fir[y]; fir[y] = e;
}

int num[maxn];
void dfs(int pos, int f) {
	int y, z;
	for (y = fir[pos]; y; y = nxt[y]) {
		if((z = to[y]) == f) continue;
		dfs(z, pos);
		num[pos] ^= num[z];
	}
	if(pos > 1) {
		if(num[pos]) printf("%d %d\n", f, pos);
		else printf("%d %d\n", pos, f);
	}
}

int main() {
	scanf("%d%d", &n, &m);
	if(m & 1) {
		printf("-1\n");
		return 0;
	}
	for (int i = 1; i <= n; ++i) fa[i] = i;
	int x, y;
	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", &x, &y);
		if(find(x) == find(y)) {
			printf("%d %d\n", x, y);
			d[x] ^= 1;
		}
		else {
			fa[find(x)] = find(y);
			add(x, y);
		}
	}
	if(d[1]) num[1] = 1;
	for (int i = 2; i <= n; ++i) {
		d[i] ^= 1;
		if(d[i]) {
			num[i] = 1;
		}
	}
	dfs(1, 0);
	return 0;
}