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 <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
template <typename _A, typename _B>
using _p = pair<_A, _B>;
template <typename _A>
using _v = vector<_A>;
template <typename _A>
using _q = queue<_A>;
template <typename _A>
using _s = set<_A>;
template <typename _A>
using _st = stack<_A>;
template <typename _A>
using _d = deque<_A>;
template <typename _A>
using _pq = priority_queue<_A>;
template <typename _A, typename _C>
using _pqc = priority_queue<_A, _v<_A>, _C>;
long long int N, S;
int tmain(void) {
  cin >> N >> S;
  auto dsum = [&](long long int n) -> long long int {
    long long int ret = 0;
    while (n) ret += n % 10, n /= 10;
    return ret;
  };
  if (dsum(N) <= S) return cout << 0 << "\n", 0;
  long long int nn = N, dn = N;
  for (long long int dg = 0, pw = 1; dg < 19; dg++, pw *= 10) {
    while (dn % 10 && dsum(nn) > S) nn += pw, dn++;
    if (dsum(nn) <= S) break;
    dn /= 10;
  }
  cout << nn - N << "\n";
  return 0;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int __T;
  cin >> __T;
  while (__T--) tmain();
  return 0;
}