Problem: You are given a positive integer n. In one move, you can increase n by one (i.e. make n := n + 1). Your task is to find the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers n and s (1 ≤ n ≤ 10^{18}; 1 ≤ s ≤ 162).

Output

For each test case, print the answer: the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.

Example

Input


5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1


Output


8
0
500
2128012501878
899999999999999999

Solution (in PYTHON3):

Solution:
def f(i, s):
    #     print(arr,s)
    if (s == arr[i]):
        if (pres[-1] - pres[i + 1] != 0):
            arr[i - 1] += 1
            for i in range(i, len(arr)):
                arr[i] = 0
            return 0
        else:
            return 0
    else:
        if (s < arr[i]):
            arr[i - 1] += 1
            for i in range(i, len(arr)):
                arr[i] = 0
            return 0
        else:
            return (s - arr[i])
for _ in range(int(input())):
    n, s = list(map(int, input().split()))
    n1 = n
    arr = []
    while (n > 0):
        arr.append(n % 10)
        n = n // 10
    if(sum(arr)<=s):
        print(0)
        continue
    arr = arr[::-1]
    arr = [0] + arr
    pres = [0] * (len(arr) + 1)
    for i in range(len(arr)):
        pres[i + 1] = pres[i] + arr[i]
    i = 1
    while (s != 0):
        s = f(i, s)
        i += 1
    for i in range(n, 0, -1):
        if (arr[i] > 9):
            arr[i - 1] += 1
            arr[i] -= 10
    if(sum(arr)<=s):
        print(0)
        continue
    c = 0
    u = 0
    for i in range(len(arr) - 1, -1, -1):
        c += (arr[i] * (10 ** (len(arr) - i - 1)))
    #     print(c)
    print(c - n1)