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:
'''
Name : Abhi Kaila
User : ak2510
Problem Name : D. Decrease the Sum of Digits
Contest : Codeforces Round #667 (Div. 3)
'''

from sys import stdin

def get_int(): return int(stdin.readline())
def get_str(): return stdin.readline().strip()
def get_ints(): return map(int, stdin.readline().strip().split())
def get_array(): return list(map(int, stdin.readline().strip().split()))
def get_array2d(row): return [list(map(int, stdin.readline().strip().split())) for _ in range(row)]

T = get_int()
for _ in range(T):
    n, s = get_ints()
    dig = [int(x) for x in str(n)]
    Sum = sum(dig)
    ans = []
    if Sum <= s:
        print(0)
    else:
        lenN = len(dig)
        c = 0
        i = lenN - 1
        while Sum + c > s:
            if dig[i] + c == 0:
                ans.append(0)
                i -= 1
                continue
            Sum = Sum - dig[i]
            ans.append(10 - dig[i] - c)
            c = 1
            dig[i] = 0
            i -= 1
        ans = ans[::-1]
        lenAns = len(ans)
        s = 0
        while ans[s]==0: s+=1
        for i in range(s, lenAns):
            print(ans[i], end="")
        print()