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;
int digit_sum(long long n) {
  int total = 0;
  while (n != 0) {
    int d = n % 10;
    total += d;
    n = n / 10;
  }
  return total;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int d;
    cin >> d;
    int sum[20] = {
        0,
    };
    for (int i = 0; i < (int)s.length(); i++) {
      sum[i + 1] = sum[i] + (s[i] - '0');
    }
    if (sum[s.length()] <= d) {
      cout << "0" << endl;
      continue;
    }
    string ans = "";
    int offset = 0;
    for (int i = (int)s.length() - 1; i >= 0; i--) {
      if (ans.length() >= 1 && digit_sum(stoll(s) + stoll(ans)) <= d) {
        break;
      }
      if (s[i] == '0' && offset == 0) {
        ans = '0' + ans;
        continue;
      }
      char c = (10 - (s[i] - '0') - offset) + '0';
      offset = 1;
      ans = c + ans;
    }
    cout << ans << endl;
  }
}