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 toll(string s) {
  long long ret = 0;
  for (int i = 0; i < s.size(); i++) {
    ret *= 10;
    ret += s[i] - '0';
  }
  return ret;
}
long long sum_digits(string s) {
  long long ret = 0;
  for (char i : s) ret += i - '0';
  return ret;
}
void solve() {
  long long n, s;
  cin >> n >> s;
  string strn = to_string(n);
  if (sum_digits(strn) <= s) {
    cout << 0 << '\n';
    return;
  }
  long long ans = 2e18;
  for (int i = 0; i < strn.size(); i++) {
    string tmp = "";
    for (int j = 0; j < i; j++) tmp += strn[j];
    if (tmp.size()) {
      tmp.back()++;
    } else {
      tmp += '1';
    }
    for (int j = i; j < strn.size(); j++) tmp += '0';
    if (sum_digits(tmp) <= s) {
      ans = min(ans, toll(tmp) - n);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}