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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    vector<long long> vec;
    long long d = 0, en = 0, i, sum = 0;
    cin >> n >> s;
    long long num = n;
    vec.push_back(0);
    while (num != 0) {
      vec.push_back(num % 10);
      d++;
      num = num / 10;
    }
    reverse(vec.begin() + 1, vec.end());
    vec.push_back(-1);
    sum = 0;
    i = 0;
    while (i < vec.size()) {
      long long e = vec[i] + 1;
      if (sum + e <= s) {
        en = i;
      }
      sum += vec[i];
      i++;
    }
    num = 0;
    for (i = 0; i < vec.size() - 1; i++) {
      if (i < en)
        num = num * 10 + vec[i];
      else if (i == en)
        num = num * 10 + (vec[i] + 1);
      else
        num = num * 10;
    }
    cout << num - n << "\n";
  }
}