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 write(vector<T> &a) {
  for (auto it = a.begin(); it != a.end(); it++) cout << *it << " ";
  cout << "\n";
}
void random(long long nMin, long long nMax, long long n) {
  srand(time(0));
  for (long long i = 0; i < n; i++) {
    double var = nMin + (long long)((double)rand() / ((double)RAND_MAX + 1) *
                                    (nMax - nMin + 1));
    cout << var << " ";
  }
  cout << "\n";
}
long long findSum(long long n) {
  long long sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
void testCase() {
  long long n, sum;
  cin >> n >> sum;
  if (findSum(n) <= sum) {
    cout << 0 << "\n";
    return;
  }
  long long ans = 0;
  long long dec = 1;
  while (true) {
    long long dig = (n / dec) % 10;
    long long add = dec * ((10 - dig) % 10);
    ans += add;
    n += add;
    if (findSum(n) <= sum) {
      break;
    }
    dec *= 10;
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(15);
  long long t = 1;
  cin >> t;
  while (t--) {
    testCase();
  }
  return 0;
}