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:

#1036C. Classy Numbers
#General method for counting numbers with some non-zero digits.
T=int(input())

def solve1(n):
    if n<=10:
        return n
    ct=0
    while n>=10:
        ct+=9
        n/=10
    ct+=n
    return ct

def solve2(n):
    if n<=100:
        return n
    m=n
    ct=0
    while m>=10:
        m/=10
        ct+=1
    k=0
    if ct>0:
        k+=9*ct
    if ct>1:
        k+=ct*(ct-1)*81/2
    k+=(m-1)*(solve1(pow(10,ct)-1)+1)
    k+=solve1(n-m*pow(10,ct))+1
    return k

def solve3(n):
    if n<=1000:
        return n
    m=n
    ct=0
    while m>=10:
        m/=10
        ct+=1
    k=0
    if ct>0:
        k+=9*ct
    if ct>1:
        k+=ct*(ct-1)*81/2
    if ct>2:
        k+=ct*(ct-1)*(ct-2)*729/6
    k+=(m-1)*(solve2(pow(10,ct)-1)+1)
    k+=(solve2(n-m*pow(10,ct))+1)
    return k

for _ in range(T):
    L,R=map(int, raw_input().split())
    print solve3(R)-solve3(L-1)

exit()