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:
def f(n):
    if n == 0:
        return 1
    dp = [[[0] * 2 for j in range(4)] for z in range(len(n))]
    dp[0][3][0] = 1
    dp[0][2][0] = int(n[0]) - 1
    dp[0][2][1] = 1
    for i in range(1, len(n)):
        for j in range(4):
            if n[i] == '0':
                dp[i][j][0] += dp[i - 1][j][0]
                dp[i][j][1] += dp[i - 1][j][1]
            else:
                dp[i][j][0] += dp[i - 1][j][0] + dp[i - 1][j][1]
            if j >= 3:
                continue
            for z in range(1, 10):
                if z < int(n[i]):
                    dp[i][j][0] += dp[i - 1][j + 1][0] + dp[i - 1][j + 1][1]
                elif z == int(n[i]):
                    dp[i][j][0] += dp[i - 1][j + 1][0]
                    dp[i][j][1] += dp[i - 1][j + 1][1]
                else:
                    dp[i][j][0] += dp[i - 1][j + 1][0]

    res = 0
    for j in range(4):
        res += dp[len(n) - 1][j][0] + dp[len(n) - 1][j][1]
    return res

t = int(input())
while t:
    t -= 1
    l, r = map(int, input().split())
    print(f(str(r)) - f(str(l - 1)))