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 sober() {
  long long int n, x = 0, a = 0, b = 0, y = 0, sum = 0, k = 0, m = 0;
  string s;
  cin >> s >> sum;
  std::vector<int> v;
  for (int i = 0; i < s.length(); i++) {
    v.push_back(s[i] - '0');
    a += s[i] - '0';
  }
  reverse(begin(v), end(v));
  if (a <= sum) {
    cout << "0\n";
    return;
  }
  long long int temp = 1, ans = 0;
  for (long long int i = 0; i < v.size(); i++) {
    ans += temp * (10 - v[i]);
    a -= v[i] - 1;
    if (i == v.size() - 1) {
      v.push_back(1);
    } else {
      v[i + 1]++;
    }
    if (a <= sum) break;
    temp *= 10;
  }
  cout << ans << "\n";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    sober();
  }
}