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>
#pragma GCC optimize("O3")
using namespace std;
const long long N = 5000005;
const long long MOD = 1e9 + 7;
long long mul(long long a, long long b) { return (a * b) % MOD; }
long long binpow(long long x, long long y) {
  long long res = 1;
  if (y <= 0) return res;
  while (y) {
    if (y & 1) res = mul(x, res);
    x = mul(x, x);
    y >>= 1;
  }
  return res;
}
long long a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m, t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    if (a % b == 0)
      cout << 0 << '\n';
    else
      cout << b - a % b << '\n';
  }
}