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;
long long d[25];
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, s;
    cin >> n >> s;
    long long len = 0, sum = 0, num = n;
    fill(d, d + 25, 0);
    while (num > 0) {
      d[++len] = num % 10;
      sum += d[len];
      num /= 10;
    }
    if (sum <= s) {
      cout << 0 << endl;
      continue;
    }
    long long cur = 0;
    for (int i = len; i >= 1; i--) {
      cur += d[i];
      if (cur >= s) {
        d[i + 1]++;
        for (int j = i; j >= 1; j--) d[j] = 0;
        break;
      }
    }
    long long ans = 0;
    cur = 1;
    for (int i = 1; i <= len + 1; i++) {
      d[i + 1] += d[i] / 10;
      d[i] = d[i] % 10;
      ans += d[i] * cur;
      cur *= 10;
    }
    cout << ans - n << endl;
  }
  return 0;
}