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>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#pragma GCC optimization("unroll-loops")
using namespace std;
using ll = long long;
const int mod = (int)1000000007;
void solve() {
  ll n, s;
  cin >> n >> s;
  deque<int> num, res;
  ll dup = n;
  while (dup) {
    num.emplace_front(dup % 10);
    dup /= 10;
  }
  res.resize(num.size());
  int tot = 0, idx = -1;
  while (tot < s) tot += num[++idx];
  bool carry = false;
  bool first = true;
  if (accumulate(num.begin(), num.end(), 0) <= s) {
    cout << 0 << endl;
    return;
  }
  while (idx >= 0) {
    if (first) {
      first = false;
      res[idx] = 0;
      carry = true;
      idx--;
      continue;
    }
    if (carry) {
      res[idx] = num[idx];
      res[idx]++;
      carry = res[idx] / 10;
      res[idx] %= 10;
    } else {
      res[idx] = num[idx];
      carry = res[idx] / 10;
    }
    idx--;
  }
  if (carry) res.push_front(1);
  string str;
  for (auto i : res) str += (i + '0');
  cout << stoull(str) - n << endl;
}
int main(int argc, char *argv[]) {
  clock_t clk_start = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int cases = 1;
  cin >> cases;
  for (int i = 1; i <= cases; i++) {
    solve();
  }
  cerr << "Execution Time: " << fixed
       << double(clock() - clk_start) / double(CLOCKS_PER_SEC)
       << setprecision(6) << " sec\n";
}