#include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long sum(long long n) {
  long long ans = 0;
  while (n > 0) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
void solve() {
  long long ans = 0, n, s, pw = 1;
  cin >> n >> s;
  if (sum(n) <= s) {
    cout << "0\n";
    return;
  }
  for (int i = 0; i < 18; i++) {
    long long cur = (n / pw) % 10;
    long long ost = pw * ((10 - cur) % 10);
    n += ost;
    ans += ost;
    if (sum(n) <= s) {
      cout << ans << "\n";
      return;
    }
    pw *= 10;
  }
}
int main() {
  long long t = 1;
  cin >> t;
  while (t--) solve();
  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;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long sum(long long n) {
  long long ans = 0;
  while (n > 0) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
void solve() {
  long long ans = 0, n, s, pw = 1;
  cin >> n >> s;
  if (sum(n) <= s) {
    cout << "0\n";
    return;
  }
  for (int i = 0; i < 18; i++) {
    long long cur = (n / pw) % 10;
    long long ost = pw * ((10 - cur) % 10);
    n += ost;
    ans += ost;
    if (sum(n) <= s) {
      cout << ans << "\n";
      return;
    }
    pw *= 10;
  }
}
int main() {
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```