#include <bits/stdc++.h>
using namespace std;
int lim, sum, pos;
long long n, res;
void calc(string s) {
  n = 0;
  pos = -1;
  for (int i = 0; i < s.size(); ++i) {
    int di = s[i] - 48;
    n = n * 10 + di;
    sum += di;
    if (sum >= lim && pos == -1) pos = i;
  }
}
long long conv(string x) {
  long long ans = 0;
  for (int i = 0; i < x.size(); ++i) {
    int di = x[i] - 48;
    ans = ans * 10 + di;
  }
  return ans;
}
long long solve(string s) {
  long long ans = conv(s.substr(0, pos)) + 1;
  int temp = s.size() - pos;
  ans *= pow(10, temp);
  if (ans == 999999999999999999) ans += 1;
  return ans - n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test;
  cin >> test;
  while (test--) {
    string s;
    sum = 0;
    cin >> s >> lim;
    calc(s);
    if (sum <= lim)
      cout << 0 << '\n';
    else
      cout << solve(s) << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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 lim, sum, pos;
long long n, res;
void calc(string s) {
  n = 0;
  pos = -1;
  for (int i = 0; i < s.size(); ++i) {
    int di = s[i] - 48;
    n = n * 10 + di;
    sum += di;
    if (sum >= lim && pos == -1) pos = i;
  }
}
long long conv(string x) {
  long long ans = 0;
  for (int i = 0; i < x.size(); ++i) {
    int di = x[i] - 48;
    ans = ans * 10 + di;
  }
  return ans;
}
long long solve(string s) {
  long long ans = conv(s.substr(0, pos)) + 1;
  int temp = s.size() - pos;
  ans *= pow(10, temp);
  if (ans == 999999999999999999) ans += 1;
  return ans - n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test;
  cin >> test;
  while (test--) {
    string s;
    sum = 0;
    cin >> s >> lim;
    calc(s);
    if (sum <= lim)
      cout << 0 << '\n';
    else
      cout << solve(s) << '\n';
  }
  return 0;
}

```