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;
template <class T, class T1>
int chkmin(T &x, const T1 &y) {
  return x > y ? x = y, 1 : 0;
}
template <class T, class T1>
int chkmax(T &x, const T1 &y) {
  return x < y ? x = y, 1 : 0;
}
const int MAXN = (1 << 20);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  cin >> T;
  for (long long t = 0; t < T; t++) {
    string s;
    long long sum;
    cin >> s >> sum;
    long long c = 0, check = 0;
    for (long long i = 0; i < s.size(); ++i) {
      check = check + s[i] - '0';
    }
    long long i = 0, summ = sum;
    for (i = 0; i < s.size(); i++) {
      if ((s[i] - '0') >= summ) {
        break;
      } else {
        summ = summ - (s[i] - '0');
      }
    }
    if (check <= sum)
      cout << 0 << '\n';
    else {
      unsigned long long xs = 1, num = 0;
      for (long long x = 0; x < s.size() - i; x++) xs = xs * 10;
      for (long long x = i; x < s.size(); x++) {
        num = num * 10 + s[x] - '0';
      }
      cout << xs - num << '\n';
    }
  }
  return 0;
}