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 main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, s, sum = 0, sz = 0;
    cin >> n >> s;
    long long tmp = n;
    vector<int> a;
    while (tmp > 0) {
      a.push_back(tmp % 10);
      sum += a.back();
      tmp /= 10;
      sz++;
    }
    reverse(a.begin(), a.end());
    if (sum <= s) {
      cout << "0\n";
    } else {
      long long temp = 0, str;
      for (int i = 0; i < a.size(); i++) {
        if (temp < s) {
          temp += a[i];
          str = i;
        } else {
          break;
        }
      }
      long long no = 0, x = 1;
      for (long long j = str; j < a.size(); j++) {
        x *= 10;
        no *= 10;
        no += a[j];
      }
      cout << x - no << "\n";
    }
  }
  return 0;
}