#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << "[ " << *it << " = " << a << " ] ";
  err(++it, args...);
}
void test(int T) {
  long long int n, s, i, j, k, p = 0, sum = 0;
  int dig[20] = {0};
  scanf("%lld %lld", &n, &s);
  for (i = 18; i >= 0; i--) {
    if (n == 0) break;
    dig[i] = n % 10;
    sum += dig[i];
    n /= 10;
  }
  j = 1;
  for (i = 18; i >= 1; i--) {
    if (sum <= s) {
      cout << p << endl;
      return;
    }
    sum -= dig[i];
    p += ((10 - dig[i]) * j);
    dig[i] = 0;
    dig[i - 1]++;
    sum++;
    j *= 10;
  }
  cout << p << endl;
}
void Test() {
  int T;
  scanf("%d\n", &T);
  for (int cs = 1; cs <= T; cs++) test(cs);
}
int main() {
  Test();
  return 0;
}

### Prompt

Please create a solution in Cpp 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 err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << "[ " << *it << " = " << a << " ] ";
  err(++it, args...);
}
void test(int T) {
  long long int n, s, i, j, k, p = 0, sum = 0;
  int dig[20] = {0};
  scanf("%lld %lld", &n, &s);
  for (i = 18; i >= 0; i--) {
    if (n == 0) break;
    dig[i] = n % 10;
    sum += dig[i];
    n /= 10;
  }
  j = 1;
  for (i = 18; i >= 1; i--) {
    if (sum <= s) {
      cout << p << endl;
      return;
    }
    sum -= dig[i];
    p += ((10 - dig[i]) * j);
    dig[i] = 0;
    dig[i - 1]++;
    sum++;
    j *= 10;
  }
  cout << p << endl;
}
void Test() {
  int T;
  scanf("%d\n", &T);
  for (int cs = 1; cs <= T; cs++) test(cs);
}
int main() {
  Test();
  return 0;
}

```