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 PYTHON):

Solution:
import math
def fun1(n):
    if n<10:
        return n
    si=len(str(n))
    return (si-1)*9+int(str(n)[0])
def fun2(n):
    if n<10:
        return 0
    ans=0
    si=len(str(n))
    if(si>=3):
        ans+=math.factorial(si-1)/(math.factorial(2)*math.factorial(si-3))*81
    ans+=(int(str(n)[0])-1)*fun1(int("9"*(si-1)))
    ans+=fun1(int(str(n)[1:]))
    return ans
def fun3(n):
    if n<100:
        return 0
    ans=0
    si=len(str(n))
    if(si>=4):
        ans+=math.factorial(si-1)/(math.factorial(3)*math.factorial(si-4))*729
    ans+=(int(str(n)[0])-1)* fun2(int("9"*(si-1)))
    ans+=fun2(int(str(n)[1:]))
    return ans
    
def fun(n):
    if len(str(n))<=3:
        return n
    ans=0
    ans+=fun1(n)
    ans+=fun2(n)
    ans+=fun3(n)
    return ans
t=int(raw_input())
for i in range(t):
    l,r=[int(i) for i in raw_input().split()]
    print fun(r)-fun(l-1)