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 long long N = (int)2e5 + 74, OO = 0x3f3f3f3f, MOD = (int)1e9 + 7;
class ADivisibilityProblem {
 public:
  void init() {}
  void solve(istream& cin, ostream& cout) {
    init();
    int T, testCases = 1;
    cin >> T;
    while (T--) {
      long long a, b;
      cin >> a >> b;
      long long d = a / b;
      if (a % b == 0)
        cout << 0 << '\n';
      else
        cout << b - (a - d * b) << '\n';
    }
  }
};
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  ADivisibilityProblem solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}