#include <bits/stdc++.h>
using namespace std;
int carry(vector<int> &digits) {
  int car = 1, idx = 0;
  while (!digits[idx]) idx++;
  digits[idx] = 0;
  for (int i = idx + 1; i < digits.size(); i++) {
    int nd = (digits[i] + car) % 10;
    car = (digits[i] + car) / 10;
    digits[i] = nd;
  }
  if (car) digits.push_back(car);
  return accumulate(digits.begin(), digits.end(), 0);
}
int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    long long n;
    int s;
    cin >> n >> s;
    vector<int> digits;
    long long n1 = n;
    while (n1) {
      digits.push_back(n1 % 10), n1 /= 10;
    }
    int s1 = accumulate(digits.begin(), digits.end(), 0);
    while (s1 > s) {
      s1 = carry(digits);
    }
    long long n2 = 0, f = 1;
    for (int i = 0; i < digits.size(); i++) {
      n2 += f * digits[i], f *= 10;
    }
    cout << n2 - n << endl;
  }
  return 0;
}

### 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;
int carry(vector<int> &digits) {
  int car = 1, idx = 0;
  while (!digits[idx]) idx++;
  digits[idx] = 0;
  for (int i = idx + 1; i < digits.size(); i++) {
    int nd = (digits[i] + car) % 10;
    car = (digits[i] + car) / 10;
    digits[i] = nd;
  }
  if (car) digits.push_back(car);
  return accumulate(digits.begin(), digits.end(), 0);
}
int main() {
  int T;
  cin >> T;
  for (int t = 1; t <= T; t++) {
    long long n;
    int s;
    cin >> n >> s;
    vector<int> digits;
    long long n1 = n;
    while (n1) {
      digits.push_back(n1 % 10), n1 /= 10;
    }
    int s1 = accumulate(digits.begin(), digits.end(), 0);
    while (s1 > s) {
      s1 = carry(digits);
    }
    long long n2 = 0, f = 1;
    for (int i = 0; i < digits.size(); i++) {
      n2 += f * digits[i], f *= 10;
    }
    cout << n2 - n << endl;
  }
  return 0;
}

```