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:
from sys import stdin

digs = [[1,0,0,0],[1,9,0,0]]

for x in range(18):
    last = digs[-1]
    temp = last[:]

    temp[1] += last[0]*9
    temp[2] += last[1]*9
    temp[3] += last[2]*9

    digs.append(temp)

def classyNums(num):
    s = str(num)

    total = 0
    things = 0
    for i,x in enumerate(s):
        nX = int(x)
        if nX:
            ind = len(s)-i

            total += sum(digs[ind-1])*nX

            total -= digs[ind-1][3]*(nX-1)
            if things > 0:
                total -= digs[ind-1][2]*(nX-1)
                total -= digs[ind-1][3]
            if things > 1:
                total -= digs[ind-1][1]*(nX-1)
                total -= digs[ind-1][2]
            if things > 2:
                total -= digs[ind-1][0]*(nX-1)
                total -= digs[ind-1][1]
            if things > 3:
                total -= digs[ind-1][0]
            things += 1
    if s.count('0') + 3 >= len(s):
        total += 1
    return total

for case in range(int(stdin.readline())):
    l,r = [int(x) for x in stdin.readline().split()]

    print(classyNums(r)-classyNums(l-1))