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(0);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    long long sum;
    cin >> sum;
    long long sum1 = 0;
    for (int i = 0; i < s.size(); i++) {
      int x = s[i] - '0';
      sum1 += x;
    }
    string a = s;
    if (sum1 > sum) {
      for (int i = s.size() - 1; i >= 0; i--) {
        int x = s[i] - '0';
        if (sum1 > sum) {
          sum1 -= x;
          s[i] = '0';
        } else {
          while (i >= 0) {
            long long y = s[i] - '0';
            sum1 -= y;
            y++;
            if (y > 9) {
              s[i] = '0';
              i--;
              continue;
            }
            sum1 += y;
            if (sum1 > sum) {
              s[i] = '0';
              sum1 -= y;
            } else {
              s[i] = y + '0';
              break;
            }
            i--;
          }
          break;
        }
      }
    } else {
      cout << 0 << "\n";
      continue;
    }
    if (s[0] == '0') s = '1' + s;
    long long p = stoll(a);
    long long q = stoll(s);
    cout << q - p << "\n";
  }
  return 0;
}