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;
template <typename T>
void print_vector(vector<T> &vec) {
  for (int i = 0; i < vec.size(); i++) {
    if (i != vec.size() - 1)
      cout << vec[i] << " ";
    else
      cout << vec[i];
  }
  cout << "\n";
}
unsigned long long int sum_digits(unsigned long long int N) {
  unsigned long long int sum = 0;
  while (N > 0) {
    sum = sum + (N % 10);
    N = N / 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  for (int t = 0; t < T; t++) {
    unsigned long long int N, S;
    cin >> N >> S;
    if (sum_digits(N) <= S) {
      cout << "0\n";
      continue;
    }
    unsigned long long int req = 0;
    unsigned long long int multi = 1;
    while (true) {
      unsigned long long int d = (N / multi) % 10;
      N = N + ((10 - d) % 10) * multi;
      req = req + ((10 - d) % 10) * multi;
      multi = multi * 10;
      if (sum_digits(N) <= S) {
        cout << req << "\n";
        break;
      }
    }
  }
}