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;
static bool comp(pair<long long int, long long int> a,
                 pair<long long int, long long int> b) {
  return a.second < b.second;
}
void read(vector<long long int> &arr, long long int n) {
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
  }
}
void write(vector<int> &arr) {
  for (auto i : arr) {
    cout << i << " ";
  }
}
void solve() {
  int t;
  cin >> t;
  for (int qq = 0; qq < t; qq++) {
    long long int n, s;
    cin >> n >> s;
    vector<int> ret(64);
    long long int k = n;
    for (int i = 63; i >= 0; i--) {
      ret[i] = k % 10;
      k /= 10;
    }
    for (int i = 63; i >= 1; i--) {
      int sum = 0;
      for (int j = 63; j >= 0; j--) {
        sum += ret[j];
      }
      if (sum <= s) {
        break;
      } else {
        ret[i] = 0;
        ret[i - 1] += 1;
      }
    }
    long long int ans = 0;
    for (int i = 0; i < 64; i++) {
      ans *= 10;
      ans += ret[i];
    }
    cout << ans - n << endl;
  }
}
void sol() {
  int n;
  cin >> n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}