#include <bits/stdc++.h>
using namespace std;
void solve();
long long toint(deque<long long> a) {
  long long ans = 0;
  for (int i = 0; i < a.size(); i++) {
    ans *= 10;
    ans += a[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}
void solve() {
  long long n, s;
  cin >> n >> s;
  deque<long long> v;
  while (n) {
    v.push_front(n % 10);
    n /= 10;
  }
  deque<long long> b(v.begin(), v.end());
  int l = v.size() - 1;
  for (int i = l; i >= 0; i--) {
    if (accumulate(v.begin(), v.end(), 0) <= s)
      goto l;
    else {
      v[i] = 0;
      if (i != 0)
        v[i - 1]++;
      else
        v.push_front(1);
    }
  }
l:
  if (v.size() == b.size() + 1) b.push_front(0);
  cout << toint(v) - toint(b);
}

### Prompt

Generate 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;
void solve();
long long toint(deque<long long> a) {
  long long ans = 0;
  for (int i = 0; i < a.size(); i++) {
    ans *= 10;
    ans += a[i];
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}
void solve() {
  long long n, s;
  cin >> n >> s;
  deque<long long> v;
  while (n) {
    v.push_front(n % 10);
    n /= 10;
  }
  deque<long long> b(v.begin(), v.end());
  int l = v.size() - 1;
  for (int i = l; i >= 0; i--) {
    if (accumulate(v.begin(), v.end(), 0) <= s)
      goto l;
    else {
      v[i] = 0;
      if (i != 0)
        v[i - 1]++;
      else
        v.push_front(1);
    }
  }
l:
  if (v.size() == b.size() + 1) b.push_front(0);
  cout << toint(v) - toint(b);
}

```