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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0);
long long getDigSum(long long x) {
  long long sum = 0;
  while (x) {
    sum += (x % 10);
    x /= 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long n_ = n, i = 1;
    while (getDigSum(n_) > s) {
      long long of = pow(10, i);
      n_ += (of - (n_ % of));
      i++;
    }
    cout << n_ - n << endl;
  }
  return 0;
}