#include <bits/stdc++.h>
using namespace std;
long long min(long long a, long long b) { return a > b ? b : a; }
long long max(long long a, long long b) { return a < b ? b : a; }
inline long long nxt() {
  long long a;
  cin >> a;
  return a;
}
inline long long digsum(long long n) {
  long long digs = 0;
  while (n != 0) {
    digs += n % 10;
    n /= 10;
  }
  return digs;
}
int main() {
  long long t = nxt();
  while (t--) {
    long long n = nxt();
    long long s = nxt();
    long long cnt = 0;
    long long itr = 1;
    while (digsum(n) > s) {
      cnt += (10 - n % 10) * itr;
      n += (10 - n % 10);
      n /= 10;
      itr *= 10;
    }
    cout << cnt << "\n";
  }
}

### Prompt

Your challenge is to write 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;
long long min(long long a, long long b) { return a > b ? b : a; }
long long max(long long a, long long b) { return a < b ? b : a; }
inline long long nxt() {
  long long a;
  cin >> a;
  return a;
}
inline long long digsum(long long n) {
  long long digs = 0;
  while (n != 0) {
    digs += n % 10;
    n /= 10;
  }
  return digs;
}
int main() {
  long long t = nxt();
  while (t--) {
    long long n = nxt();
    long long s = nxt();
    long long cnt = 0;
    long long itr = 1;
    while (digsum(n) > s) {
      cnt += (10 - n % 10) * itr;
      n += (10 - n % 10);
      n /= 10;
      itr *= 10;
    }
    cout << cnt << "\n";
  }
}

```