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;
char _;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  int t;
  cin >> t;
  while (t--) {
    unsigned long long n;
    int s;
    cin >> n >> s;
    string num = to_string(n);
    int keep = 0;
    int sum = 0;
    for (int i = 0; i < num.length(); i++) {
      int digit = num[i] - '0';
      if ((i == num.length() - 1 && sum + digit == s) || sum + digit < s) {
        keep++;
        sum += digit;
      } else {
        break;
      }
    }
    sum = 0;
    for (int i = 0; i < num.length(); i++) {
      sum += num[i] - '0';
    }
    string ans;
    if (sum <= s) {
      cout << 0 << endl;
      continue;
    }
    if (keep == 0) {
      ans = "1";
      for (int i = 0; i < num.length(); i++) {
        ans += '0';
      }
    } else {
      ans = to_string(stoull(num.substr(0, keep)) + 1);
      while (ans.length() < num.length()) {
        ans += '0';
      }
    }
    long long newVal = stoull(ans);
    if (newVal < n) {
      newVal = stoull(to_string(newVal) + "0");
    }
    cout << newVal - n << endl;
  }
  return 0;
}