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).
import heapq as hq
from heapq import heappop,heappush
from collections import deque,defaultdict,Counter


def inp():
	return (int(input()))


def inlt():
	return (list(map(int,input().split())))


def insr():
	s=input()
	return (list(s[:len(s)]))


def invr():
	return (map(int,input().split()))


def subset_sum_count(arr,n,sum):
	dp=[[0 for _ in range(sum+1)] for _ in range(n+1)]
	for i in range(n+1):
		for j in range(sum+1):
			if j==0:
				dp[i][j]=1
			elif arr[i-1]<=j:
				dp[i][j]=dp[i-1][j-arr[i-1]]+dp[i-1][j]
			else:
				dp[i][j]=dp[i-1][j]
	return dp[n][sum]


def prefix(a):
	pre=[]
	pre.append(a[0])
	for i in range(1,len(a)):
		pre.append(pre[i-1]+a[i])
	return pre


def binary_search(func,lo,hi,abs_prec=1e-7):
	""" Locate the first value x s.t. func(x) = True within [lo, hi] """
	while abs(hi-lo)>abs_prec:
		mi=lo+(hi-lo)/2
		if func(mi):
			hi=mi
		else:
			lo=mi
	
	return (lo+hi)/2


def ternary_search(func,lo,hi,abs_prec=1e-7):
	""" Find maximum of unimodal function func() within [lo, hi] """
	while abs(hi-lo)>abs_prec:
		lo_third=lo+(hi-lo)/3
		hi_third=hi-(hi-lo)/3
		
		if func(lo_third)<func(hi_third):
			lo=lo_third
		else:
			hi=hi_third
	
	return (lo+hi)/2


def discrete_binary_search(func,lo,hi):
	""" Locate the first value x s.t. func(x) = True within [lo, hi] """
	while lo<hi:
		mi=lo+(hi-lo)//2
		if func(mi):
			hi=mi
		else:
			lo=mi+1
	
	return lo


def discrete_ternary_search(func,lo,hi):
	""" Find the first maximum of unimodal function func() within [lo, hi] """
	while lo<=hi:
		lo_third=lo+(hi-lo)//3
		hi_third=lo+(hi-lo)//3+(1 if 0<hi-lo<3 else (hi-lo)//3)
		
		if func(lo_third)<func(hi_third):
			lo=lo_third+1
		else:
			hi=hi_third-1
	
	return lo
	
	return -1


def right_rotate(a,s):
	return a[s:]+a[:s]


def dec_to_bin(x):
	return int(bin(x)[2:])


def str_to_integer_list(n):
	a=[]
	for i in range(len(n)):
		a.append(int(n[i]))
	return a


def list_to_str(l):
	s=""
	for i in l:
		s+=str(i)
	return s


def dijkstra(s,N,E):
	visited=set()
	dist={}
	for i in range(1,N+1):
		dist[i]=1<<29
	queue=[(dist[i],i) for i in range(1,N+1)]
	hq.heappush(queue,(0,s))
	dist[s]=0
	while queue:
		d,u=hq.heappop(queue)
		if u in visited:
			continue
		#Relax all the neighbours of u
		for t in E[u]:
			v,r=t
			if dist[v]>d+r:
				dist[v]=d+r
				hq.heappush(queue,(dist[v],v))
		#Node u has been processed
		visited.add(u)
	return dist


def prime_sieve(n):
	"""returns a sieve of primes >= 5 and < n"""
	flag=n%6==2
	sieve=bytearray((n//3+flag>>3)+1)
	for i in range(1,int(n**0.5)//3+1):
		if not (sieve[i>>3]>>(i&7))&1:
			k=(3*i+1)|1
			for j in range(k*k//3,n//3+flag,2*k):
				sieve[j>>3]|=1<<(j&7)
			for j in range(k*(k-2*(i&1)+4)//3,n//3+flag,2*k):
				sieve[j>>3]|=1<<(j&7)
	return sieve


def prime_list(n):
	"""returns a list of primes <= n"""
	res=[]
	if n>1:
		res.append(2)
	if n>2:
		res.append(3)
	if n>4:
		sieve=prime_sieve(n+1)
		res.extend(3*i+1|1 for i in range(1,(n+1)//3+(n%6==1)) if not (sieve[i>>3]>>(i&7))&1)
	return res


def dijkstra(n,graph,start):
	""" Uses Dijkstra's algortihm to find the shortest path between in a graph. """
	dist,parents=[float("inf")]*n,[-1]*n
	dist[start]=0
	queue=[(0,start)]
	while queue:
		path_len,v=heappop(queue)
		if path_len==dist[v]:
			for w,edge_len in graph[v]:
				if edge_len+path_len<dist[w]:
					dist[w],parents[w]=edge_len+path_len,v
					heappush(queue,(edge_len+path_len,w))
	
	return dist,parents


def path(start,end,parent):
	path=[end]
	while path[-1]!=start:
		path.append(parent[path[-1]])
	path.reverse()
	return path


def bfs(graph,start,goal):
	"""
	finds a shortest path in undirected `graph` between `start` and `goal`.
	If no path is found, returns `None`
	"""
	if start==goal:
		return [start]
	visited={start}
	queue=deque([(start,[])])
	
	while queue:
		current,path=queue.popleft()
		visited.add(current)
		for neighbor in graph[current]:
			if neighbor==goal:
				return path+[current,neighbor]
			if neighbor in visited:
				continue
			queue.append((neighbor,path+[current]))
			visited.add(neighbor)
	return None  # no path found. not strictly needed


'''for _ in range(int(input())):
	n,m=list(map(int,input().split()))
	graph=defaultdict(list)
	distance=[-1]*(n+1)
	visited=[False]*(n+1)
	for _ in range(m):
		x,y=list(map(int,input().split()))
		graph[x].append(y)
		graph[y].append(x)'''
'''for _ in range(int(input())):
	n,m=list(map(int,input().split()))
	graph=[[]for _ in range(n+1)]
	distance=[-1]*(n+1)
	visited=[False]*(n+1)
	for _ in range(m):
		x,y,w=list(map(int,input().split()))
		graph[x].append([y,w])
		graph[y].append([x,w])'''

import os
import sys
from io import BytesIO,IOBase
def DecimalToBinary(n):

    def helper(n):
        if n == 0:
            return '' # <- empty string
        else:
            if n%2 == 1:
                binaryDigit = '1'
            else:
                binaryDigit = '0'
            return helper(n//2) + binaryDigit

    if n == 0:
        result = '0' # <- "leading zero" only for `n == 0`
    else:
        result = helper(n)
    return result

def p(board):
	for i in range(8):
		s=""
		for j in range(8):
			s+=board[i][j]
		print(s)
def binaryToDecimal(n):
    return int(n,2)
def decimalToBinary(n):
    return bin(n).replace("0b","")
def solve(a,b):
	x=DecimalToBinary(a);y=DecimalToBinary(b)
	z=x+y
	l=y+x
	
	r= binaryToDecimal(z)-binaryToDecimal(l)
	return r


def mostFrequent(arr,n):
	# Sort the array
	arr.sort()
	
	# find the max frequency using
	# linear traversal
	max_count=1;
	res=arr[0];
	curr_count=1
	
	for i in range(1,n):
		if (arr[i]==arr[i-1]):
			curr_count+=1
		
		else:
			if (curr_count>max_count):
				max_count=curr_count
				res=arr[i-1]
			
			curr_count=1
	
	# If last element is most frequent
	if (curr_count>max_count):
		max_count=curr_count
		res=arr[n-1]
	
	return res
from bisect import bisect_right,bisect_left


def lds(arr,n):
	lds=[0]*n
	max=0
	
	for i in range(n):
		lds[i]=1
	
	for i in range(1,n):
		for j in range(i):
			if (arr[i]<arr[j] and
					lds[i]<lds[j]+1):
				lds[i]=lds[j]+1
	
	for i in range(n):
		if (max<lds[i]):
			max=lds[i]
	
	# returns the length of the LDS
	return max


import math


def solve():
	for _ in range(inp()):
		
		a,b=invr()
		x=abs(b-a)
		if x%10==0:
			print(x//10)
		else:
			print((x//10)+1)
			
		
		


solve()
				
			
	
	
		
		
		
