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;
const int inf = 0x3f3f3f3f;
bool ispalindrome(string s) {
  long long int len = s.length();
  for (long long int i = 0; i < len; i++) {
    if (s[i] != s[len - i - 1]) return false;
  }
  return true;
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res) * (a));
    a = ((a) * (a));
    b >>= 1;
  }
  return res;
}
long long is_prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long int sum(long long int n) {
  long long int add = 0;
  while (n) {
    add += (n % 10);
    n /= 10;
  }
  return add;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 0;
  cin >> t;
  for (long long int test = 0; test < t; test++) {
    long long int n, s;
    cin >> n >> s;
    if (sum(n) <= s)
      cout << 0 << '\n';
    else {
      long long int p = 1;
      long long int m = n;
      long long int cn = 0;
      while (1) {
        int dig = (m / p) % 10;
        long long int add = p * ((10 - dig) % 10);
        n += add;
        m = n;
        cn += add;
        if (sum(n) <= s) break;
        p *= 10;
      }
      cout << cn << '\n';
    }
  }
  return 0;
}