#include <bits/stdc++.h>
using namespace std;
#define N 200005
#define whyqxxx
#define M N

int n, m;
int tot = 0, fir[N], nxt[N << 1], to[N << 1];
int x[N << 1], y[N << 1], z[N << 1];

inline void addEdge(int u, int v) {
    nxt[++tot] = fir[u], fir[u] = tot, to[tot] = v;
}

int vis[N], d[N];

void dfs(int u) {
    vis[u] = 1;
    for (int i = fir[u], v; i; i = nxt[i]) {
        v = to[i]; if (vis[v]) continue;
        dfs(v);
        if (d[v]) z[(i + 1) / 2] ^= 1, d[v] = 0, d[u] ^= 1;
    }
}

int main() {
#ifdef whyqx
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif
    cin >> n >> m;
    for (int i = 1; i <= m; ++i)
        scanf("%d%d", x + i, y + i),
        addEdge(x[i], y[i]), addEdge(y[i], x[i]), d[x[i]] ^= 1, z[i] = 1;
    dfs(1);
    if (d[1]) {
        puts("-1");
        return 0;
    }
    for (int i = 1; i <= m; ++i)
        if (z[i]) printf("%d %d\n", x[i], y[i]);
        else printf("%d %d\n", y[i], x[i]);
    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>
using namespace std;
#define N 200005
#define whyqxxx
#define M N

int n, m;
int tot = 0, fir[N], nxt[N << 1], to[N << 1];
int x[N << 1], y[N << 1], z[N << 1];

inline void addEdge(int u, int v) {
    nxt[++tot] = fir[u], fir[u] = tot, to[tot] = v;
}

int vis[N], d[N];

void dfs(int u) {
    vis[u] = 1;
    for (int i = fir[u], v; i; i = nxt[i]) {
        v = to[i]; if (vis[v]) continue;
        dfs(v);
        if (d[v]) z[(i + 1) / 2] ^= 1, d[v] = 0, d[u] ^= 1;
    }
}

int main() {
#ifdef whyqx
    freopen("work.in", "r", stdin);
    freopen("work.out", "w", stdout);
#endif
    cin >> n >> m;
    for (int i = 1; i <= m; ++i)
        scanf("%d%d", x + i, y + i),
        addEdge(x[i], y[i]), addEdge(y[i], x[i]), d[x[i]] ^= 1, z[i] = 1;
    dfs(1);
    if (d[1]) {
        puts("-1");
        return 0;
    }
    for (int i = 1; i <= m; ++i)
        if (z[i]) printf("%d %d\n", x[i], y[i]);
        else printf("%d %d\n", y[i], x[i]);
    return 0;
}

```