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;
using ll = long long;
const int INF = 1000000000;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(12);
  }
} iosetup;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++)
    os << v[i] << (i + 1 == (int)v.size() ? "" : " ");
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}
using int128 = __int128;
ostream &operator<<(ostream &os, const __int128 n) {
  if (n == 0) {
    os << 0;
    return os;
  }
  __int128 num = n;
  bool neg = false;
  if (num < 0) neg = true, num = -num;
  string res = "";
  while (num > 0) {
    res.push_back('0' + num % 10);
    num /= 10;
  }
  if (neg) res.push_back('-');
  reverse(begin(res), end(res));
  os << res;
  return os;
}
istream &operator>>(istream &is, __int128 &n) {
  string s;
  is >> s;
  int idx = 0;
  bool neg = false;
  if (s[0] == '-') neg = true, idx++;
  n = 0;
  for (; idx < (int)s.size(); idx++) {
    n = 10 * n + s[idx] - '0';
  }
  if (neg) n = -n;
  return is;
}
__int128 abs128(__int128 x) { return x < 0 ? -x : x; }
__int128 gcd128(__int128 a, __int128 b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return b > 0 ? gcd128(b, a % b) : a;
}
int f(int128 x) {
  int ret = 0;
  while (x) {
    ret += x % 10;
    x /= 10;
  }
  return ret;
}
void solve() {
  int128 n;
  int s;
  cin >> n >> s;
  int128 cost = 1, res = 0;
  while (f(n) > s) {
    res += cost * (10 - n % 10);
    n /= 10;
    n++;
    cost *= 10;
  }
  cout << res << endl;
}
signed main() {
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}