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;
void solve() {
  string n;
  int s;
  cin >> n >> s;
  int sum = 0;
  vector<int> digit;
  for (int i = 0; i < (int)n.size(); i++) {
    digit.push_back(n[i] - '0');
    sum += digit[i];
  }
  if (sum <= s) {
    cout << 0 << "\n";
    return;
  }
  sum = 0;
  int pos = -1;
  for (int i = 0; i < (int)digit.size(); i++) {
    sum += digit[i];
    if (sum >= s) {
      pos = i;
      s -= (sum - digit[i]);
      break;
    }
  }
  if (pos == 0) {
    digit.insert(digit.begin(), 1);
    pos++;
  } else {
    digit[pos - 1]++;
  }
  for (int i = pos; i < (int)digit.size(); i++) digit[i] = 0;
  long long num = 0;
  for (int i = 0; i < (int)digit.size(); i++) num = num * 10 + digit[i];
  long long res = num - stoll(n);
  cout << res << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}