human
What is a Python code to print any one of them given the following:
Can you solve the below in PYTHON3?
This is the easy version of the problem. The only difference between the two versions is the constraint on n. 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 1 to n, 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) is added to Mocha's forest, then an edge (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 n, m_1 and m_2 (1 ≤ n ≤ 1000, 0 ≤ m_1, m_2 < n) — the number of nodes and the number of initial edges in Mocha's forest and Diana's forest.

Each of the next m_1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) — the edges in Mocha's forest.

Each of the next m_2 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) — the edges in Diana's forest.

Output

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

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

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

Examples

Input

3 2 2
1 2
2 3
1 2
1 3

Output

0

Input

5 3 2
5 4
2 1
4 3
4 3
1 4

Output

1
2 4

Input

8 1 2
1 7
2 6
1 5

Output

5
5 2
2 3
3 4
4 7
6 8

Note

In the first example, we cannot add any edge.

In the second example, the initial forests are as follows.


We can add an edge (2, 4).
assistant
#Here is Python code to print any one of them.

# ///////////////////////////////////////////////////////////////////////////
# //////////////////// PYTHON IS THE BEST ////////////////////////
# ///////////////////////////////////////////////////////////////////////////
import sys,os,io
from sys import stdin
import math 
from collections import defaultdict
from heapq import heappush, heappop, heapify
from bisect import bisect_left , bisect_right
from io import BytesIO, IOBase
BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")



alphabets = list('abcdefghijklmnopqrstuvwxyz')


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