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;
const long long M = 1e9 + 7;
void solve() {
  string n;
  long long int s;
  cin >> n >> s;
  int z = n.size();
  int arr[z];
  int sum = 0;
  for (int i = 0; i < z; i++) {
    int p = n[i] - '0';
    arr[i] = p;
    sum += p;
  }
  if (sum <= s) {
    cout << 0 << endl;
  } else {
    sum = 0;
    int flag = 0;
    for (int i = 0; i < z; i++) {
      sum += arr[i];
      if (sum >= s) {
        flag = i;
        break;
      }
    }
    long long int g = 1;
    long long int h = 0;
    for (int i = z - 1; i >= flag; i--) {
      h += g * arr[i];
      g *= 10;
    }
    cout << g - h << endl;
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}