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 nax = 1000 * 1007;
const int INF = 1e9;
long long int pow(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int nod(long long int n) {
  long long int co = 0;
  while (n > 0) {
    co++;
    n = n / 10;
  }
  return co;
}
long long int find(long long int n, long long int s) {
  long long int x = nod(n);
  long long int ne = 0, n1 = n;
  long long int s1 = s;
  for (int i = x; i >= 1; i--) {
    long long int calc = pow(10, i - 1);
    long long int d = n / calc;
    s = s - d;
    if (s < 0) {
      ne = ne + pow(10, i);
      break;
    }
    ne = ne + d * calc;
    n = n % calc;
  }
  if (ne == n1) return ne;
  return find(ne, s1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long int n, s;
    cin >> n >> s;
    long long int x = find(n, s);
    cout << (x - n) << endl;
  }
}