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;

typedef long long ll;
typedef unsigned long long ull;
struct E {
  int v;
  bool vis;
  E *next, *rev;
};

const int N = 100010;

int n, m;
bool vis[N], par[N];
E* g[N];

E* adde(int u, int v) {
  static E pool[N * 2], *p = pool;
  ++p;
  p->v = v;
  p->next = g[u];
  g[u] = p;
  return p;
}

void dfs(int u) {
  vis[u] = true;
  for (E* p = g[u]; p; p = p->next)
    if (!p->vis) {
      p->vis = p->rev->vis = true;
      if (!vis[p->v]) {
        dfs(p->v);
        if (par[p->v]) {
          printf("%d %d\n", p->v, u);
        } else {
          printf("%d %d\n", u, p->v);
          par[u] ^= 1;
        }
      } else {
        printf("%d %d\n", u, p->v);
        par[u] ^= 1;
      }
    }
}

int main() {
  scanf("%d%d", &n, &m);
  if (m % 2) {
    puts("-1");
    return 0;
  }
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    E *p = adde(u, v), *q = adde(v, u);
    p->rev = q;
    q->rev = p;
  }
  dfs(1);
  return 0;
}