You are given two integers a and b.

In one move, you can choose some integer k from 1 to 10 and add it to a or subtract it from a. In other words, you choose an integer k ∈ [1; 10] and perform a := a + k or a := a - k. You may use different values of k in different moves.

Your task is to find the minimum number of moves required to obtain b from a.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers a and b (1 ≤ a, b ≤ 10^9).

Output

For each test case, print the answer: the minimum number of moves required to obtain b from a.

Example

Input


6
5 5
13 42
18 4
1337 420
123456789 1000000000
100500 9000


Output


0
3
2
92
87654322
9150

Note

In the first test case of the example, you don't need to do anything.

In the second test case of the example, the following sequence of moves can be applied: 13 → 23 → 32 → 42 (add 10, add 9, add 10).

In the third test case of the example, the following sequence of moves can be applied: 18 → 10 → 4 (subtract 8, subtract 6).


# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
# sys.setrecursionlimit(10**5)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]











def solution(a,b):
	if b == a:
		return 0
	if b > a and (b-a)%2 == 1:
		return 1
	if b < a and (b-a)%2 == 0:
		return 1
	return 2






def main():
	T = read_int()
	for i in range(T):
		a,b = read_int_array()
		x = solution(a,b)

		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()



'''stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* BITS - THINK HOW TO MASK PROPERLY
	* PERMUTATIONS - PARITY AND CYCLES
	* Think simple, if it becomes over complicated, try to look at it from a different perspective.
	* Have fun!!!
	* TRY FIXING SOMETHING, and then maybe binary search around it.
	* Remember heaps. 
	* Remember how to add a value to a segment when using prefix sum.
		suppose you have an array[1,2,3,4,5] and you want to add 3 to array[1:4]. Then just 
		add 3 to A[1], and decrease 3 from A[4]. Let's look at what happens:
		original prefixsums is 					 [1,3,6,10,15]
		array -> [1,5,3,4,2] and prefix sums are [1,6,9,13,15] 
		As you see, exactly +3 in A[1:4]
		*** The previous method can help checking how many x,y you can choose to get s=x+y from two arrays.
'''
'''
binary search 
	while(r - l > 1) {
		ll mid = l + (r - l) / 2;
		solve(mid);
		ll sum = 0;
		for (int i = 0; i < n; i++)
			sum += b[i];
		if (sum <= k)
			r = mid;
		else
			l = mid;
	}
'''
