#include <bits/stdc++.h>
#define x first
#define y second

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

int n, m;
vector<int> lis[100100];
int vis[100100];
int col[100100];

int tc = 1;
void dfs(int here, int p) {
    vis[here] = tc++;
    for (int &there : lis[here]) {
        if (!vis[there]) {
            dfs(there,here);
        }
        else if (there!=p&&vis[there]<vis[here]) {
            printf("%d %d\n",here+1,there+1);
            col[here] ^= 1;
        }
    }
    if (~p) {
        if (col[here]) {
            printf("%d %d\n",here+1,p+1);
        }
        else {
            printf("%d %d\n",p+1,here+1);
            col[p]^=1;
        }
    }
}

int main() {
    scanf("%d%d",&n,&m);
    for (int i=0;i<m;i++) {
        int a, b;
        scanf("%d%d",&a,&b);
        a--; b--;
        lis[a].push_back(b);
        lis[b].push_back(a);
    }
    if (m%2) puts("-1");
    else {
        dfs(0,-1);
    }

    return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 <bits/stdc++.h>
#define x first
#define y second

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

int n, m;
vector<int> lis[100100];
int vis[100100];
int col[100100];

int tc = 1;
void dfs(int here, int p) {
    vis[here] = tc++;
    for (int &there : lis[here]) {
        if (!vis[there]) {
            dfs(there,here);
        }
        else if (there!=p&&vis[there]<vis[here]) {
            printf("%d %d\n",here+1,there+1);
            col[here] ^= 1;
        }
    }
    if (~p) {
        if (col[here]) {
            printf("%d %d\n",here+1,p+1);
        }
        else {
            printf("%d %d\n",p+1,here+1);
            col[p]^=1;
        }
    }
}

int main() {
    scanf("%d%d",&n,&m);
    for (int i=0;i<m;i++) {
        int a, b;
        scanf("%d%d",&a,&b);
        a--; b--;
        lis[a].push_back(b);
        lis[b].push_back(a);
    }
    if (m%2) puts("-1");
    else {
        dfs(0,-1);
    }

    return 0;
}

```