#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, a, b, x, y, n, s;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    x = n, y = 0;
    int i = 0;
    int a[20];
    while (x > 0) {
      a[i] = x % 10;
      y += x % 10;
      x /= 10;
      i++;
    }
    if (y <= s) {
      cout << 0 << "\n";
    } else {
      long long d = y - s;
      d++;
      long long f = i;
      for (int j = 0; j < i; j++) {
        if (d > a[j]) {
          d = d - a[j];
          a[j] = 0;
        } else {
          a[j] = 0;
          f = j;
          break;
        }
      }
      int r = 1;
      for (int j = f + 1; j < i; j++) {
        a[j] = a[j] + r;
        r = a[j] / 10;
        a[j] = a[j] % 10;
      }
      if (r != 0) {
        a[i] = 1;
        i++;
      }
      long long p = 0;
      for (int j = i - 1; j >= 0; j--) {
        p = p * 10 + a[j];
      }
      cout << p - n << "\n";
    }
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, a, b, x, y, n, s;
  cin >> t;
  while (t--) {
    cin >> n >> s;
    x = n, y = 0;
    int i = 0;
    int a[20];
    while (x > 0) {
      a[i] = x % 10;
      y += x % 10;
      x /= 10;
      i++;
    }
    if (y <= s) {
      cout << 0 << "\n";
    } else {
      long long d = y - s;
      d++;
      long long f = i;
      for (int j = 0; j < i; j++) {
        if (d > a[j]) {
          d = d - a[j];
          a[j] = 0;
        } else {
          a[j] = 0;
          f = j;
          break;
        }
      }
      int r = 1;
      for (int j = f + 1; j < i; j++) {
        a[j] = a[j] + r;
        r = a[j] / 10;
        a[j] = a[j] % 10;
      }
      if (r != 0) {
        a[i] = 1;
        i++;
      }
      long long p = 0;
      for (int j = i - 1; j >= 0; j--) {
        p = p * 10 + a[j];
      }
      cout << p - n << "\n";
    }
  }
  return 0;
}

```