#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const long long N = 1e3 + 10;
const long long IM = 1e15 + 37;
const long double PI = 3.1415926535897932384;
inline void PP(long long n) { cout << n << " "; }
void PV(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
signed main() {
  op();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    string n;
    long long s;
    cin >> n >> s;
    long long x = 0;
    for (long long i = 0; i < n.size(); i++) {
      x += n[i] - '0';
    }
    if (x <= s) {
      cout << (0) << "\n";
      continue;
    }
    reverse(n.begin(), n.end());
    n += "000";
    long long cn = 0, r = 0;
    long long tr = 1;
    for (long long i = 0; i < n.size(); i++) {
      long long u = n[i] - '0';
      u = 10 - u;
      cn += u * tr;
      tr *= 10;
      long long ck = 0;
      r = 1;
      for (long long j = i + 1; j < n.size(); j++) {
        if (n[j] == '9' && r) {
          n[j] = '0';
        } else if (r) {
          n[j]++;
          r = 0;
        }
        ck += n[j] - '0';
      }
      if (ck <= s) break;
    }
    cout << (cn) << "\n";
  }
  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;
const long long M = 1e9 + 7;
const long long N = 1e3 + 10;
const long long IM = 1e15 + 37;
const long double PI = 3.1415926535897932384;
inline void PP(long long n) { cout << n << " "; }
void PV(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
signed main() {
  op();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    string n;
    long long s;
    cin >> n >> s;
    long long x = 0;
    for (long long i = 0; i < n.size(); i++) {
      x += n[i] - '0';
    }
    if (x <= s) {
      cout << (0) << "\n";
      continue;
    }
    reverse(n.begin(), n.end());
    n += "000";
    long long cn = 0, r = 0;
    long long tr = 1;
    for (long long i = 0; i < n.size(); i++) {
      long long u = n[i] - '0';
      u = 10 - u;
      cn += u * tr;
      tr *= 10;
      long long ck = 0;
      r = 1;
      for (long long j = i + 1; j < n.size(); j++) {
        if (n[j] == '9' && r) {
          n[j] = '0';
        } else if (r) {
          n[j]++;
          r = 0;
        }
        ck += n[j] - '0';
      }
      if (ck <= s) break;
    }
    cout << (cn) << "\n";
  }
  return 0;
}

```