#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long a[21] = {0};
    long long p = 20;
    while (n > 0) {
      a[p--] = n % 10;
      n /= 10;
    }
    long long sum = 0;
    for (long long i = 0; i <= 20; i++) sum += a[i];
    if (sum <= s)
      cout << 0 << endl;
    else {
      long long val = 0;
      long long carry = 0;
      long long hold = 1;
      for (int i = 20; i >= 1; i--) {
        sum -= (a[i]);
        val += (10 - a[i]) * hold;
        a[i] = 0;
        carry = 1;
        hold *= 10;
        for (int j = i - 1; j >= 0; j--) {
          long long v = a[j];
          a[j] = (a[j] + carry) % 10;
          carry = (v + carry) / 10;
        }
        long long chk = 0;
        for (int j = 0; j <= 20; j++) chk += a[j];
        if (chk <= s) {
          cout << val << endl;
          break;
        }
      }
    }
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long a[21] = {0};
    long long p = 20;
    while (n > 0) {
      a[p--] = n % 10;
      n /= 10;
    }
    long long sum = 0;
    for (long long i = 0; i <= 20; i++) sum += a[i];
    if (sum <= s)
      cout << 0 << endl;
    else {
      long long val = 0;
      long long carry = 0;
      long long hold = 1;
      for (int i = 20; i >= 1; i--) {
        sum -= (a[i]);
        val += (10 - a[i]) * hold;
        a[i] = 0;
        carry = 1;
        hold *= 10;
        for (int j = i - 1; j >= 0; j--) {
          long long v = a[j];
          a[j] = (a[j] + carry) % 10;
          carry = (v + carry) / 10;
        }
        long long chk = 0;
        for (int j = 0; j <= 20; j++) chk += a[j];
        if (chk <= s) {
          cout << val << endl;
          break;
        }
      }
    }
  }
  return 0;
}

```