You are given a square matrix consisting of n rows and n columns. We assume that the rows are numbered from 1 to n from top to bottom and the columns are numbered from 1 to n from left to right. Some cells (n - 1 cells in total) of the the matrix are filled with ones, the remaining cells are filled with zeros. We can apply the following operations to the matrix:

  1. Swap i-th and j-th rows of the matrix; 
  2. Swap i-th and j-th columns of the matrix. 



You are asked to transform the matrix into a special form using these operations. In that special form all the ones must be in the cells that lie below the main diagonal. Cell of the matrix, which is located on the intersection of the i-th row and of the j-th column, lies below the main diagonal if i > j.

Input

The first line contains an integer n (2 ≤ n ≤ 1000) — the number of rows and columns. Then follow n - 1 lines that contain one's positions, one per line. Each position is described by two integers xk, yk (1 ≤ xk, yk ≤ n), separated by a space. A pair (xk, yk) means that the cell, which is located on the intersection of the xk-th row and of the yk-th column, contains one.

It is guaranteed that all positions are distinct.

Output

Print the description of your actions. These actions should transform the matrix to the described special form.

In the first line you should print a non-negative integer m (m ≤ 105) — the number of actions. In each of the next m lines print three space-separated integers t, i, j (1 ≤ t ≤ 2, 1 ≤ i, j ≤ n, i ≠ j), where t = 1 if you want to swap rows, t = 2 if you want to swap columns, and i and j denote the numbers of rows or columns respectively.

Please note, that you do not need to minimize the number of operations, but their number should not exceed 105. If there are several solutions, you may print any of them.

Examples

Input

2
1 2


Output

2
2 1 2
1 1 2


Input

3
3 1
1 3


Output

3
2 2 3
1 1 3
1 1 2


Input

3
2 1
3 2


Output

0

Difficulty: HARD
Tags: Greedy algorithms

Mathematics

Constructive algorithms
Skill types: Greedy algorithms

Starter code:

Solutions:
import os
from io import BytesIO, IOBase
import sys

def main():
	n = int(input())
	row = [0] * n
	col = [0] * n
	a = [[0 for j in range(n)] for i in range(n)]
	for i in range(n - 1):
		(x, y) = map(int, input().split())
		row[x - 1] += 1
		col[y - 1] += 1
		a[x - 1][y - 1] = 1
	ans = []
	for i in range(n):
		zci = -1
		for j in range(n - i):
			if col[j] == 0:
				zci = j
				break
		if zci == -1:
			break
		if zci != n - i - 1:
			ans.append([2, zci + 1, n - i])
			(col[n - i - 1], col[zci]) = (col[zci], col[n - i - 1])
			for j in range(n - i):
				(a[j][zci], a[j][n - i - 1]) = (a[j][n - i - 1], a[j][zci])
		nzri = -1
		for j in range(n - i):
			if row[j] != 0:
				nzri = j
				break
		if nzri == -1:
			break
		if nzri != n - i - 1:
			ans.append([1, nzri + 1, n - i])
			(row[nzri], row[n - i - 1]) = (row[n - i - 1], row[nzri])
		for j in range(n - i):
			if a[nzri][j] == 1:
				col[j] -= 1
			(a[nzri][j], a[n - i - 1][j]) = (a[n - i - 1][j], a[nzri][j])
	print(len(ans))
	for i in ans:
		print(*i)
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')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')
main()


def find_all(s, c):
	index = s.index(c)
	while True:
		yield index
		try:
			index = s.index(c, index + 1)
		except ValueError:
			break
n = int(input())
(row, col, actions) = ([], [], [])
available = list(range(1, n + 1))
for _ in range(n - 1):
	(r, c) = map(int, input().split(' '))
	row.append(r)
	col.append(c)
for cur in range(n, 1, -1):
	if not row:
		break
	if cur in col:
		indices = list(find_all(col, cur))
		zero = list(set(available) - set(col))[0]
		actions.append([2, cur, zero])
		for i in indices:
			col[i] = zero
	try:
		idx = row.index(cur)
		indices = list(find_all(row, cur))
		indices.reverse()
		for i in indices:
			del row[i], col[i]
		del available[-1]
		continue
	except ValueError:
		val = row[0]
		indices = list(find_all(row, val))
		actions.append([1, row[0], cur])
		indices.reverse()
		for i in indices:
			del row[i], col[i]
		del available[-1]
if not actions:
	print(0)
else:
	print(len(actions))
	for move in actions:
		print(*move)


n = int(input())
P = [list(map(int, input().split())) for _ in range(n - 1)]
col = set()
row = set()
sCol = {}
sRow = {}
for (x, y) in P:
	row.add(x)
	col.add(y)
n_col = len(col)
n_row = len(row)
(cur_col, cur_row) = (1, n)
used_col = {}
used_row = {}
for c in sorted(col):
	if c > n_col:
		while cur_col in used_col:
			cur_col += 1
		used_col[cur_col] = 1
		sCol[c] = cur_col
	else:
		used_col[c] = 1
for r in sorted(row, reverse=True):
	if r < n - n_row + 1:
		while cur_row in used_row:
			cur_row -= 1
		used_row[cur_row] = 1
		sRow[r] = cur_row
	else:
		used_row[r] = 1
for (i, [x, y]) in enumerate(P):
	if x in sRow:
		P[i][0] = sRow[x]
	if y in sCol:
		P[i][1] = sCol[y]
minCol = [float('inf') for _ in range(n_col + 1)]
minCol[0] = 0
for (x, y) in P:
	if x < minCol[y]:
		minCol[y] = x

def sort(minCol):
	swap = []
	pos = {}
	for (col, val) in enumerate(minCol):
		if col == 0:
			continue
		if val not in pos:
			pos[val] = set()
		pos[val].add(col)
	sortCol = sorted(minCol)
	for (i, val) in enumerate(sortCol):
		if i == 0:
			continue
		if val == minCol[i]:
			pos[val].remove(i)
		else:
			col_ = next(iter(pos[val]))
			pos[val].remove(col_)
			pos[minCol[i]].remove(i)
			pos[minCol[i]].add(col_)
			minCol[col_] = minCol[i]
			minCol[i] = val
			swap.append([2, i, col_])
	return swap
swap = sort(minCol)
ans = [[1, x, y] for (x, y) in sRow.items()] + [[2, x, y] for (x, y) in sCol.items()] + swap
print(len(ans))
print('\n'.join([' '.join([str(x) for x in line]) for line in ans]))


n = int(input())
a = []
b = [0] * (n + 1)
c = [10000] * (n + 1)
for i in range(0, n - 1):
	(p1, p2) = list(map(int, input().split()))
	a.append((p1, p2))
	b[p1] = max(b[p1], p2)
ans = []
for i in range(1, n + 1):
	if b[i] == 0:
		continue
	k = 0
	for j in range(i, n + 1):
		if b[j] == 0:
			k = j
			break
	if k == 0:
		break
	b[j] = b[i]
	b[i] = 0
	for j in range(0, n - 1):
		if a[j][0] == i:
			a[j] = (k, a[j][1])
	ans.append((1, i, k))
for i in a:
	c[i[1]] = min(c[i[1]], i[0])
for i in range(1, n + 1):
	k = i
	for j in range(i + 1, n + 1):
		if c[j] < c[i]:
			i = j
	if i == k:
		continue
	ans.append((2, i, k))
	c[0] = c[i]
	c[i] = c[k]
	c[k] = c[0]
print(len(ans))
for i in ans:
	print(i[0], i[1], i[2])


n = int(input())
(a, b, c) = ([], [0] * (n + 1), [int(2000000000.0)] * (n + 1))
for i in range(n - 1):
	(p1, p2) = map(int, input().split())
	a.append((p1, p2))
	b[p1] = max(b[p1], p2)
res = []
for i in range(1, n + 1):
	if b[i] == 0:
		continue
	k = 0
	for j in range(i, n + 1):
		if b[j] == 0:
			k = j
			break
	else:
		break
	(b[j], b[i]) = (b[i], 0)
	for j in range(n - 1):
		if a[j][0] == i:
			a[j] = (k, a[j][1])
	res.append((1, i, k))
for i in a:
	c[i[1]] = min(c[i[1]], i[0])
for i in range(1, n + 1):
	k = i
	for j in range(i + 1, n + 1):
		if c[j] < c[i]:
			i = j
	if i == k:
		continue
	res.append((2, i, k))
	(c[0], c[i], c[k]) = (c[i], c[k], c[0])
print(len(res))
for i in res:
	print(i[0], i[1], i[2])