Problem: Let's call some positive integer classy if its decimal representation contains no more than 3 non-zero digits. For example, numbers 4, 200000, 10203 are classy and numbers 4231, 102306, 7277420000 are not.

You are given a segment [L; R]. Count the number of classy integers x such that L ≤ x ≤ R.

Each testcase contains several segments, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^4) — the number of segments in a testcase.

Each of the next T lines contains two integers L_i and R_i (1 ≤ L_i ≤ R_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of classy integers on a segment [L_i; R_i].

Example

Input

4
1 1000
1024 1024
65536 65536
999999 1000001


Output

1000
1
0
2

Solution (in PYTHON3):

Solution:
import sys
from sys import stdin,stdout
def ncr(n,r):
	if n<r or r<0 or n<0:
		return 0
	if n-r<r:
		r=n-r
	arr=[0 for i in range(r+1)]
	arr[0]=1
	for i in range(n+1):
		for j in range(min(i,r),0,-1):
			arr[j]=arr[j]+arr[j-1]
	return arr[r]
##print(ncr(10,5))
def getclassy(s):
	#print("\t\t",s)
	ans=0;nzd=0
	for i in range(len(s)):
		if s[i]=='0':
			continue
		#ASSUMING ZERO DIGIT
		#print([ncr(len(s)-i-1,j)*9**j for j in range(3-nzd+1)])
		ans+=(sum([ncr(len(s)-i-1,j)*9**j for j in range(3-nzd+1)]));nzd+=1
		if nzd==4:
			break;
		ans+=(sum([ncr(len(s)-i-1,j)*9**j for j in range(3-nzd+1)]))*(int(s[i])-1)
		#print(ans,"ANS",i)
	return ans

t=int(stdin.readline())
for _ in range(t):
	l,r=map(int,stdin.readline().split(' '));r+=1;l=list(str(l));r=list(str(r))
	stdout.write(str(getclassy(r)-getclassy(l))+"\n")