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 mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long ans, total;
    long long i = 0;
    string s;
    long long init = 0;
    long long curr = 0;
    cin >> s >> total;
    long long j = s.length() - 1;
    long long xx = 10;
    for (i = 0; i < s.length(); i++) init = xx * init + (s[i] - '0');
    for (i = 0; i < s.length(); i++) curr += (s[i] - '0');
    while (curr > total) {
      s[j] = '0';
      curr = 0;
      for (i = j - 1; i >= 0; i--) {
        if (s[i] == '9')
          s[i] = '0';
        else {
          s[i] = s[i] + 1;
          break;
        }
      }
      if (s[0] == '0') s = '1' + s;
      for (long long kk = 0; kk < s.length(); kk++) curr = curr + (s[kk] - '0');
      j--;
    }
    long long out = 0;
    for (long long kk = 0; kk < s.length(); kk++)
      out = out * 10ll + (s[kk] - '0');
    cout << out - init << "\n";
  }
  return 0;
}