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 *
from copy import *
from string import *				# alpha = ascii_lowercase
from random import *
from sys import stdin
from sys import maxsize
from operator import *				# d = sorted(d.items(), key=itemgetter(1))
from itertools import *
from collections import Counter		# d = dict(Counter(l))
import math

def bin1(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=k**mid
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin1(mid+1,r,k,t,b,val,ans)
		else:
			return bin1(l,mid-1,k,t,b,val,ans)
		
def bin2(l,r,k,t,b,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		v=t*(k**mid)+b*(mid)
		if(v==val):
			return v
		elif(v>val):
			ans=mid
			return bin2(l,mid-1,k,t,b,val,ans)
		else:
			return bin2(mid+1,r,k,t,b,val,ans)

def SieveOfEratosthenes(n): 
      
    # Create a boolean array "prime[0..n]" and initialize 
    #  all entries it as true. A value in prime[i] will 
    # finally be false if i is Not a prime, else true. 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        # If prime[p] is not changed, then it is a prime 
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    l=[]
    for i in range(2,n+1):
    	if(prime[i]):
    		l.append(i)
    return l
def bin(l,r,ll,val):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		if(val>=ll[mid][0] and val<=ll[mid][1]):
			return mid
		elif(val<ll[mid][0]):
			return bin(l,mid-1,ll,val)
		else:
			return bin(mid+1,r,ll,val)
def deci(n):
	s=""
	while(n!=0):
		if(n%2==0):
			n=n//2
			s="0"+s
		else:
			n=n//2
			s="1"+s
	return s
def diff(s1,s2):
	if(len(s1)<len(s2)):
		v=len(s1)
		while(v!=len(s2)):
			s1="0"+s1
			v=v+1
	else:
		v=len(s2)
		while(v!=len(s1)):
			s2="0"+s2
			v=v+1
	c=0
	for i in range(len(s1)):
		if(s1[i:i+1]!=s2[i:i+1]):
			c=c+1
	return c
from sys import stdin, stdout 
def fac(a,b):
	v=a
	while(a!=b):
		v*=a-1
		a=a-1
	return v
def bino(l,r,n):
	if(l>r):
		return -1
	else:
		mid=(l+r)//2
		val1=math.log((n/mid)+1,2)
		val2=int(val1)
		if(val1==val2):
			return val1
		elif(val1<1.0):
			return bino(l,mid-1,n)
		else:
			return bino(mid+1,r,n)

def binary(l,r,ll,val,ans):
	if(l>r):
		return ans
	else:
		mid=(l+r)//2
		if(ll[mid]<=val):
			ans=mid
			return binary(mid+1,r,ll,val,ans)
		else:
			return binary(l,mid-1,ll,val,ans)

if __name__ == '__main__':
	t=int(stdin.readline())
	for i in range(t):
		a,b=map(int,stdin.readline().split(" "))
		c=0
		if(a==b):
			print(0)
		elif(a>b):
			val=a-b
			if(val%2==0):
				print(1)
			else:
				print(2)
		else:
			val=b-a
			if(val%2==0):
				print(2)
			else:
				print(1)