#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
void run_case() {
  uint64_t N, S;
  cin >> N >> S;
  string str = to_string(N);
  str = "0" + str;
  int n = int(str.size());
  vector<uint64_t> candidates = {N};
  for (int i = n - 1; i >= 0; i--) {
    if (str[i] != '9') {
      str[i]++;
      candidates.push_back(stoull(str));
    }
    str[i] = '0';
  }
  auto &&sum_of_digits = [](uint64_t x) {
    uint64_t sum = 0;
    while (x != 0) {
      sum += x % 10;
      x /= 10;
    }
    return sum;
  };
  uint64_t best = 1e19;
  for (uint64_t cand : candidates)
    if (sum_of_digits(cand) <= S) best = min(best, cand - N);
  cout << best << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  while (tests-- > 0) run_case();
}

### Prompt

Your challenge is to write a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
void run_case() {
  uint64_t N, S;
  cin >> N >> S;
  string str = to_string(N);
  str = "0" + str;
  int n = int(str.size());
  vector<uint64_t> candidates = {N};
  for (int i = n - 1; i >= 0; i--) {
    if (str[i] != '9') {
      str[i]++;
      candidates.push_back(stoull(str));
    }
    str[i] = '0';
  }
  auto &&sum_of_digits = [](uint64_t x) {
    uint64_t sum = 0;
    while (x != 0) {
      sum += x % 10;
      x /= 10;
    }
    return sum;
  };
  uint64_t best = 1e19;
  for (uint64_t cand : candidates)
    if (sum_of_digits(cand) <= S) best = min(best, cand - N);
  cout << best << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tests;
  cin >> tests;
  while (tests-- > 0) run_case();
}

```