#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    unsigned long long int N;
    int sum;
    string s;
    cin >> N >> sum;
    s = to_string(N);
    s = "0" + s;
    int suma = 0;
    for (auto it : s) suma += (it - '0');
    int p = s.size() - 1;
    while (suma > sum && p >= 0) {
      if (s[p] == '0') {
        --p;
        continue;
      }
      s[p] = '0';
      ++s[p - 1];
      for (int i = p - 1; i >= 0; --i) {
        if (int(s[i]) == 58) {
          s[i] = '0';
          if (i > 0)
            ++s[i - 1];
          else
            s = "1" + s;
        }
      }
      suma = 0;
      for (auto it : s) suma += (it - '0');
      --p;
    }
    unsigned long long nr = 0;
    for (auto it : s) nr = nr * 10 + (it - '0');
    cout << nr - N << '\n';
  }
}

### 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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    unsigned long long int N;
    int sum;
    string s;
    cin >> N >> sum;
    s = to_string(N);
    s = "0" + s;
    int suma = 0;
    for (auto it : s) suma += (it - '0');
    int p = s.size() - 1;
    while (suma > sum && p >= 0) {
      if (s[p] == '0') {
        --p;
        continue;
      }
      s[p] = '0';
      ++s[p - 1];
      for (int i = p - 1; i >= 0; --i) {
        if (int(s[i]) == 58) {
          s[i] = '0';
          if (i > 0)
            ++s[i - 1];
          else
            s = "1" + s;
        }
      }
      suma = 0;
      for (auto it : s) suma += (it - '0');
      --p;
    }
    unsigned long long nr = 0;
    for (auto it : s) nr = nr * 10 + (it - '0');
    cout << nr - N << '\n';
  }
}

```