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


def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def LS2(): return list(sys.stdin.readline().rstrip())


def f(X):  # X以下で条件を満たす整数の個数
    digit = len(X)
    dp = [[[0]*4 for _ in range(2)] for _ in range(digit+1)]
    # pos,is_less,count_nonzero
    dp[0][0][0] = 1
    for d in range(digit):
        x = X[d]
        for i in range(2):
            for j in range(4):
                upper = 9 if i else x
                for k in range(upper+1):
                    if j+(k!=0) <= 3:
                        dp[d+1][i|(k<x)][j+(k!=0)] += dp[d][i][j]
    res0 = sum(dp[-1][0])
    res1 = sum(dp[-1][1])
    return res0,res1


t = I()
for _ in range(t):
    L,R = S().split()
    L = [int(l) for l in L]
    R = [int(r) for r in R]
    l0,l1 = f(L)
    r0,r1 = f(R)
    ans = r0+r1-l1
    print(ans)