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import sys
import bisect
def count_digit(d, n):
    if n == 0:
        return 0
    count = 0
    for i in range(10, 0, -1):
        divisor = (n + i - 1) // i
        next_divisor = divisor // 10
        current = divisor % 10
        if current == d:
            count += (n - divisor * i + 1)
        elif current < d:
            count += next_divisor * i
        if d == 0:
            count += next_divisor * i
        else:
            count += (next_divisor - (1 if current < d else 0)) * i
        if next_divisor == 0:
            break
    return count
MAX_X = 1018
# Precompute for each d, the k where f(d, k) =k, up to max_x
max_precompute = 10**5
dp = [[0] * (max_precompute + 1) for _ in range(10)]
for d in range(1, 10):
    for k in range(0, max_precompute + 1):
        dp[d][k] = count_digit(d, k)
q = int(sys.stdin.readline())
for _ in range(q):
    line = sys.stdin.readline().strip()
    if not line:
        continue
    parts = line.split()
    d = int(parts[0])
    x = int(parts[1])
    if x == 0:
        print(0)
        continue
    # Find the largest k <= x where dp[d][k] ==k
    pre = dp[d][x]
    if pre >= x:
        # More checking since dp[d][x]==pre >=x is not possible
        print(x)
    else:
        # Binary search between pre and x
        # The values are not guaranteed to be increasing?
        # Wait, actually, f(d, k) increases with k for each d, but f(d, k) could be greater than k.
        # Therefore, for some k, if dp[d][k] =k, as k increases, does the next solution can jump?
        # But in the problem's sample, the function f(d, k) could have several k's that satisfy.
        # So binary search may not find the solution, because the solution could be in a jump.
        # Alternative: collect all such k for all d, up to given x, then for each query, directly output the maximum <=x.
        # Then, search this list, as the number of such k's is small.
        # Let's try this.
        print(0)

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