import sys
import math

# Optimized version of the solve function to reduce time complexity
def solve(x, m):
    # Precompute all necessary powers of 10 to avoid redundant recalculations
    num = [1] * 40
    for i in range(1, 37):
        num[i] = num[i - 1] * 10

    # Convert x into its digit array in reverse order
    a = []
    while x > 0:
        a.append(x % 10)
        x //= 10
    cnt = len(a)

    # If x is 0, we have one digit
    if cnt == 0:
        a.append(0)
        cnt = 1

    # Base case for m == 0, we check for the first non-9 digit
    if m == 0:
        for i in range(cnt):
            if a[i] != 9:
                print(num[i])
                return

    # Use a dynamic programming array to track minimum costs
    dp = [[float('inf')] * (m + 1) for _ in range(2)]
    dp[0][0] = 0

    # Iterate over each digit position
    for i in range(cnt):
        current = i % 2
        next_ = (i + 1) % 2
        dp[next_] = [float('inf')] * (m + 1)
        
        for k in range(m + 1):
            # No carry case
            if a[i] + 1 < 10:
                dp[next_][k] = min(dp[next_][k], dp[current][k])
            else:
                dp[next_][k] = min(dp[next_][k], dp[current][k])

            # Carry case, only if we still have carries left
            if k < m:
                if a[i] != 0:
                    dp[next_][k + 1] = min(dp[next_][k + 1], (10 - a[i]) * num[i] + dp[current][k])
                else:
                    dp[next_][k + 1] = min(dp[next_][k + 1], (10 - a[i] - 1) * num[i] + dp[current][k])

    ans = float('inf')
    for i in range(m + 1):
        ans = min(ans, dp[cnt % 2][i])

    print(ans)
    return

def main():
    input = sys.stdin.read
    data = input().split()
    t = int(data[0])
    idx = 1
    results = []
    for _ in range(t):
        x = int(data[idx])
        m = int(data[idx + 1])
        idx += 2
        solve(x, m)

if __name__ == "__main__":
    main()