human
PYTHON3 solution for "It is a simplified version of problem F2. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).

You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.

A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2).

Output

For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.

Example

Input

4
1 1
221 2
177890 2
998244353 1

Output

1
221
181111
999999999":


 What is a Python program to find the minimum integer x such that x ≥ n and the number x is k-beautiful?
assistant
#Here is Python program to find the minimum integer x such that x ≥ n and the number x is k-beautiful.
 def nr_distinct(n):
	return len(set(list(n)))

def get(n, idx, k):
	if n[idx] == '9':
		return -1
	
	digs = set(list(n[:idx]))

	if len(digs) > k:
		return -1
	
	if len(digs) == k:
		x = chr(ord('9') + 1)
		y = x
		for ch in digs:
			if ord(ch) > ord(n[idx]) and ord(ch) < ord(x):
				x = ch
			if ord(ch) < ord(y):
				y = ch

		if ord(x) <= ord('9'):
			return n[:idx] + x + "".join([y] * (len(n) - idx - 1))
		else:
			return -1
	else:
		x = chr(ord(n[idx]) + 1)
		digs.add(x)

		if len(digs) < k:
			y = '0'
		else:
			y = chr(ord('9') + 1)
			for ch in digs:
				if ord(ch) < ord(y):
					y = ch

		return n[:idx] + x + "".join([y] * (len(n) - idx - 1))

for _ in range(int(input())):
	n, k = input().split()
	k = int(k)

	if nr_distinct(n) <= k:
		print(n)
		continue
	
	found = False
	for idx in range(len(n) - 1, -1, -1):
		ans = get(n, idx, k)
		if ans != -1:
			print(ans)
			found = True
			break

	if not found:
		print("".jioin(['1'] * (len(n) + 1)))