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;
void solve() {
  long long n, s;
  cin >> n >> s;
  string curr = to_string(n);
  long long currSum = 0;
  bool flag = false;
  for (long long i = 0; i < curr.size(); i++) {
    currSum += curr[i] - '0';
    if (currSum > s) {
      long long temp = i - 1;
      while (temp >= 0 && curr[temp] == '9') {
        temp--;
      }
      if (temp < 0)
        flag = true;
      else {
        curr[temp]++;
        for (long long j = temp + 1; j < curr.size(); j++) curr[j] = '0';
      }
      break;
    } else if (currSum == s) {
      bool flag1 = true;
      for (long long j = i + 1; j < curr.size(); j++)
        if (curr[j] != '0') flag1 = false;
      if (!flag1) {
        long long temp = i - 1;
        while (temp >= 0 && curr[temp] == '9') {
          temp--;
        }
        if (temp < 0)
          flag = true;
        else {
          curr[temp]++;
          for (long long j = temp + 1; j < curr.size(); j++) curr[j] = '0';
        }
      }
      break;
    }
  }
  if (flag) {
    string temp = "1";
    for (long long i = 0; i < curr.size(); i++) {
      temp += '0';
    }
    curr = temp;
  }
  long long now = 0;
  for (long long i = 0; i < curr.size(); i++) {
    long long x = curr[i] - '0';
    now = now * 10 + x;
  }
  cout << now - n << '\n';
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}