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;
long long bin(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  if (b % 2 == 0) return bin((a * a) % mod, b / 2, mod) % mod;
  return ((a % mod) * (bin((a * a) % mod, b / 2, mod) % mod)) % mod;
}
void solve() {
  string s;
  cin >> s;
  int val = 0;
  cin >> val;
  set<int> num;
  reverse(s.begin(), s.end());
  int val1 = 0;
  for (int i = 0; i < s.size(); i++) {
    val1 += s[i] - '0';
  }
  long long ans = 0;
  int n = s.size();
  for (int i = 0; i < s.size() && val1 > val; i++) {
    if (s[i] == '0') {
      continue;
    }
    long long ch = s[i] - '0';
    ans += (10 - ch) * bin(10, i, LLONG_MAX);
    val1 -= ch;
    s[i] = '0';
    int j = i + 1;
    while (j < s.size()) {
      if (s[j] == '9') {
        s[j] = '0';
        val1 -= 9;
        j++;
      } else {
        s[j]++;
        val1++;
        break;
      }
    }
    if (s[s.size()] == '0') {
      s.push_back('1');
      val1++;
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}