D1. Mocha and Diana (Easy Version)＜738,div2＞

D1. Mocha and Diana (Easy Version)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

This is the easy version of the problem. The only difference between the two versions is the constraint on nn. You can make hacks only if all versions of the problem are solved.

A forest is an undirected graph without cycles (not necessarily connected).

Mocha and Diana are friends in Zhijiang, both of them have a forest with nodes numbered from 11 to nn, and they would like to add edges to their forests such that:

• After adding edges, both of their graphs are still forests.
• They add the same edges. That is, if an edge (u,v)(u,v) is added to Mocha's forest, then an edge (u,v)(u,v) is added to Diana's forest, and vice versa.

Mocha and Diana want to know the maximum number of edges they can add, and which edges to add.

Input

The first line contains three integers nn, m1m1 and m2m2 (1≤n≤10001≤n≤1000, 0≤m1,m2<n0≤m1,m2<n) — the number of nodes and the number of initial edges in Mocha's forest and Diana's forest.

Each of the next m1m1 lines contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — the edges in Mocha's forest.

Each of the next m2m2 lines contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — the edges in Diana's forest.

Output

The first line contains only one integer hh, the maximum number of edges Mocha and Diana can add (in each forest).

Each of the next hh lines contains two integers uu and vv (1≤u,v≤n1≤u,v≤n, u≠vu≠v) — the edge you add each time.

If there are multiple correct answers, you can print any one of them.

Examples

input

Copy

3 2 2

1 2

2 3

1 2

1 3

output

Copy

0

input

Copy

5 3 2

5 4

2 1

4 3

4 3

1 4

output

Copy

1

2 4

input

Copy

8 1 2

1 7

2 6

1 5

output

Copy

5

5 2

2 3

3 4

4 7

6 8

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int>p;
int a[10010];
int n = 10005;
int father[10005], father1[10005];
void init(int n) {
  for (int i = 1; i <= n; ++i) {
    father[i] = i;
    father1[i] = i;
  }
}
int find1(int u) {//根
  return u == father1[u] ? u : father1[u] = find1(father1[u]);
}
int find(int u) {//根
  return u == father[u] ? u : father[u] = find(father[u]);
}
bool same(int u, int v) {
  u = find(u);
  v = find(v);
  return u == v;
}
int main() {
  int m1, m2;
  cin >> n >> m1 >> m2;
  init(n);//父亲节点是自己
  while (m1--) {
    int u, v;
    cin >> u >> v;
    int x = find(u), y = find(v); //树建立
    if (x != y)
      father[x] = y;
  }
  while (m2--) {
    int u, v;
    cin >> u >> v;
    int x = find1(u), y = find1(v);
    if (x != y)
      father1[x] = y;
  }
  vector<p>v;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      int x = find(i), xx = find1(i), y = find(j), yy = find1(j);
      if (x != y && yy != xx) {
        father[x] = y;
        father1[xx] = yy;
        v.push_back(p(i, j));
      }
    }
  }
  int cnt = v.size();
  cout << cnt << endl;
  for (int i = 0; i < cnt; i++) {
    cout << v[i].first << " " << v[i].second << endl;
  }
}