#include <cstdio>
#include <vector>
using namespace std;
#define pb push_back
const int maxn = 100001;
int N,M,vis[maxn]; vector<int> adj[maxn],adj2[maxn];
bool dfs(int cur,int par) {
    vis[cur] = 1;
    for (int nxt : adj[cur]) {
        if (!vis[nxt]) {
            if (dfs(nxt,cur)) adj2[nxt].pb(cur);
            else adj2[cur].pb(nxt);
        }
        else if (nxt != par && vis[nxt] == 1) adj2[cur].pb(nxt);
    }
    vis[cur] = 2;
    return (int)adj2[cur].size()&1;
}
int main() {
    scanf("%d%d",&N,&M);
    for (int i = 0,A,B;i < M;i++) scanf("%d%d",&A,&B),adj[A].pb(B),adj[B].pb(A);
    if (M&1) return printf("-1\n"),0;
    dfs(1,0); for (int i = 1;i <= N;i++) for (int j : adj2[i]) printf("%d %d\n",i,j);
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <cstdio>
#include <vector>
using namespace std;
#define pb push_back
const int maxn = 100001;
int N,M,vis[maxn]; vector<int> adj[maxn],adj2[maxn];
bool dfs(int cur,int par) {
    vis[cur] = 1;
    for (int nxt : adj[cur]) {
        if (!vis[nxt]) {
            if (dfs(nxt,cur)) adj2[nxt].pb(cur);
            else adj2[cur].pb(nxt);
        }
        else if (nxt != par && vis[nxt] == 1) adj2[cur].pb(nxt);
    }
    vis[cur] = 2;
    return (int)adj2[cur].size()&1;
}
int main() {
    scanf("%d%d",&N,&M);
    for (int i = 0,A,B;i < M;i++) scanf("%d%d",&A,&B),adj[A].pb(B),adj[B].pb(A);
    if (M&1) return printf("-1\n"),0;
    dfs(1,0); for (int i = 1;i <= N;i++) for (int j : adj2[i]) printf("%d %d\n",i,j);
}

```