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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  string st;
  int s;
  while (t--) {
    cin >> st;
    cin >> s;
    int cSum = 0;
    int l = st.size();
    for (int i = 0; i < l; i++) cSum += (st[i] - '0');
    if (cSum <= s) {
      cout << "0" << '\n';
      continue;
    }
    reverse((st).begin(), (st).end());
    long long mov = 0;
    bool ok = 1, ok2 = 1;
    st.resize(l + 5);
    for (int i = 0;; i++) {
      int d = st[i] - '0';
      if (d == 0) continue;
      long long p = pow(10, i) + 0.1;
      mov = mov + (10 - d) * p;
      st[i + 1] = (((st[i + 1] - '0') + 1) % 10) + '0';
      cSum -= (d - 1);
      if (cSum <= s) {
        cout << mov << '\n';
        ok = 0;
        break;
      }
      while (st[i + 1] == '0') {
        cSum -= 9;
        if (cSum <= s) {
          cout << mov << '\n';
          ok = 0;
          break;
        }
        i++;
        st[i + 1] = (((st[i + 1] - '0') + 1) % 10) + '0';
      }
      if (!ok) break;
    }
  }
  return 0;
}